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Binary Search Trees 2 solutions added
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‎07_Binary-Search-Tree/Binary-Search-Trees-2/Problems_Binary_Search_Tree_2.ipynb

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{
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"nbformat": 4,
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"nbformat_minor": 0,
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"metadata": {
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"colab": {
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"name": "Problems_Binary-Search-Tree-2",
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"provenance": [],
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"collapsed_sections": []
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},
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"kernelspec": {
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"name": "python3",
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"display_name": "Python 3"
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},
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"language_info": {
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"name": "python"
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}
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},
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"cells": [
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{
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"cell_type": "markdown",
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"source": [
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"**CONTENT**"
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],
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"metadata": {
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"id": "o7_kl1tdQBpY"
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}
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},
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{
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"cell_type": "code",
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"source": [
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"class BinaryTreeNode:\n",
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"\n",
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" def __init__(self,data):\n",
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" self.data=data;\n",
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" self.left=None\n",
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" self.right=None\n",
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" \n",
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"def search(root,x):\n",
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" if root==None:\n",
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" return False\n",
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" if root.data==x:\n",
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" return True\n",
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" elif root.data>x:\n",
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" return search(root.left,x)\n",
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" else:\n",
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" return search(root.left,x)\n",
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"\n",
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"def printTreeDetailed(root):\n",
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" if root==None:\n",
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" return\n",
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" print(root.data,end=\":\")\n",
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" if root.left!=None:\n",
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" print(\"L\",root.left.data,end=\",\")\n",
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" if root.right!=None:\n",
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" print(\"R\",root.right.data,end=\"\")\n",
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" print()\n",
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" printTreeDetailed(root.left)\n",
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" printTreeDetailed(root.right)\n",
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"\n",
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"import queue\n",
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"def takeTreeInputLevelWise():\n",
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" q=queue.Queue()\n",
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" print(\"Enter root\")\n",
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" rootData=int(input())\n",
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" if rootData==-1:\n",
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" return None\n",
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" root=BinaryTreeNode(rootData)\n",
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" q.put(root)\n",
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" while(not(q.empty())):\n",
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" current_node=q.get()\n",
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" print(\"Enter left child of\",current_node.data)\n",
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" leftChildData=int(input())\n",
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" if leftChildData!=-1:\n",
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" leftChild=BinaryTreeNode(leftChildData)\n",
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" current_node.left=leftChild\n",
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" q.put(leftChild)\n",
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" print(\"Enter right child of\",current_node.data)\n",
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" rightChildData=int(input())\n",
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" if rightChildData!=-1:\n",
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" rightChild=BinaryTreeNode(rightChildData)\n",
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" current_node.right=rightChild\n",
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" q.put(rightChild)\n",
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" return root\n",
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"\n",
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"def nodeToRootPath(root,s):\n",
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" if root==None:\n",
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" return None\n",
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" if root.data==s:\n",
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" l=list()\n",
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" l.append(root.data)\n",
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" return l\n",
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" leftOutput=nodeToRootPath(root.left,s)\n",
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" if leftOutput!=None:\n",
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" leftOutput.append(root.data)\n",
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" return leftOutput\n",
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" rightOutput=nodeToRootPath(root.right,s)\n",
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" if rightOutput!=None:\n",
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" rightOutput.append(root.data)\n",
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" return rightOutput\n",
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" else:\n",
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" return None\n",
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"\n",
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"root=takeTreeInputLevelWise()\n",
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"printTreeDetailed(root)\n",
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"nodeToRootPath(root,5)"
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],
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"metadata": {
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"id": "ne3Yz9XAQECO"
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},
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"execution_count": null,
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"outputs": []
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},
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{
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"cell_type": "markdown",
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"source": [
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"**CONTENT PROBLEMS**"
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],
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"metadata": {
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"id": "LAMbOztNJgpj"
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}
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {
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"id": "7hl6_i-8JW-_"
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},
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"outputs": [],
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"source": [
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"'''\n",
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"Find path in BST\n",
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"\n",
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"Given a BST and an integer k. Find and return the path from the node with data k and root \n",
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"(if a node with data k is present in given BST) in a list. Return empty list otherwise.\n",
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"\n",
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"Note: Assume that BST contains all unique elements.\n",
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"\n",
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"Input Format :\n",
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"The first line of input contains data of the nodes of the tree in level order form. \n",
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"The data of the nodes of the tree is separated by space. If any node does not have left or right child, \n",
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"take -1 in its place. Since -1 is used as an indication whether the left or right nodes exist, therefore, \n",
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"it will not be a part of the data of any node. \n",
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"\n",
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"The following line of input contains an integer, that denotes the value of k.\n",
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"\n",
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"Output Format :\n",
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"The first line and only line of output prints the data of the nodes in the path from node k to root. \n",
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"The data of the nodes is separated by single space.\n",
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"\n",
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"Constraints:\n",
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"Time Limit: 1 second \n",
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"\n",
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"Sample Input 1:\n",
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"8 5 10 2 6 -1 -1 -1 -1 -1 7 -1 -1\n",
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"2\n",
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"Sample Output 1:\n",
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"2 5 8\n",
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"'''\n",
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"\n",
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"import queue\n",
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"class BinaryTreeNode:\n",
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" def __init__(self, data):\n",
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" self.data = data\n",
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" self.left = None\n",
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" self.right = None\n",
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"\n",
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"def findPathBST(root,data):\n",
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" if root is None:\n",
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" return None\n",
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" if root.data == data:\n",
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" lst = list()\n",
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" lst.append(root.data)\n",
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" return lst\n",
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" leftOutput = findPathBST(root.left,data)\n",
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" if leftOutput != None:\n",
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" leftOutput.append(root.data)\n",
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" return leftOutput\n",
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" rightOutput = findPathBST(root.right,data)\n",
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" if rightOutput != None:\n",
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" rightOutput.append(root.data)\n",
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" return rightOutput\n",
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" \n",
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"def buildLevelTree(levelorder):\n",
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" index = 0\n",
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" length = len(levelorder)\n",
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" if length<=0 or levelorder[0]==-1:\n",
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" return None\n",
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" root = BinaryTreeNode(levelorder[index])\n",
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" index += 1\n",
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" q = queue.Queue()\n",
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" q.put(root)\n",
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" while not q.empty():\n",
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" currentNode = q.get()\n",
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" leftChild = levelorder[index]\n",
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" index += 1\n",
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" if leftChild != -1:\n",
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" leftNode = BinaryTreeNode(leftChild)\n",
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" currentNode.left =leftNode\n",
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" q.put(leftNode)\n",
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" rightChild = levelorder[index]\n",
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" index += 1\n",
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" if rightChild != -1:\n",
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" rightNode = BinaryTreeNode(rightChild)\n",
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" currentNode.right =rightNode\n",
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" q.put(rightNode)\n",
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" return root\n",
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"\n",
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"# Main\n",
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"levelOrder = [int(i) for i in input().strip().split()]\n",
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"root = buildLevelTree(levelOrder)\n",
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"data = int(input())\n",
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"path = findPathBST(root,data)\n",
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"if path is not None:\n",
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" for ele in path:\n",
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" print(ele,end=' ')"
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]
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},
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{
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"cell_type": "code",
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"source": [
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"'''\n",
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"BST Class\n",
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"\n",
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"Implement the BST class which includes following functions -\n",
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"\n",
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"1. search\n",
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"Given an element, find if that is present in BST or not. Return true or false.\n",
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"\n",
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"2. insert -\n",
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"Given an element, insert that element in the BST at the correct position. \n",
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"If element is equal to the data of the node, insert it in the left subtree.\n",
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"\n",
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"3. delete -\n",
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"Given an element, remove that element from the BST. \n",
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"If the element which is to be deleted has both children, \n",
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"replace that with the minimum element from right sub-tree.\n",
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"\n",
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"4. printTree (recursive) -\n",
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"Print the BST in ithe following format -\n",
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"\n",
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"For printing a node with data N, you need to follow the exact format -\n",
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"N:L:x,R:y\n",
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"\n",
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"where, N is data of any node present in the binary tree. \n",
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"x and y are the values of left and right child of node N. Print the children only if it is not null.\n",
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"\n",
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"There is no space in between.\n",
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"You need to print all nodes in the recursive format in different lines.\n",
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"\n",
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"'''\n",
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"\n",
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"\n",
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"\n",
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"class BinaryTreeNode:\n",
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" def __init__(self, data):\n",
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" self.data = data\n",
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" self.left = None\n",
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" self.right = None\n",
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" \n",
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"class BST:\n",
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" \n",
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" def __init__(self):\n",
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" self.root = None\n",
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" self.numNodes = 0\n",
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" \n",
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" \n",
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" def printTreeHelper(self,root):\n",
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" if root==None:\n",
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" return\n",
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" print(root.data,end=\":\")\n",
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" if root.left!=None:\n",
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" print(\"L:\" + str(root.left.data),end=\",\")\n",
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" if root.right!=None:\n",
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" print(\"R:\" + str(root.right.data),end=\"\")\n",
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" print()\n",
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" self.printTreeHelper(root.left)\n",
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" self.printTreeHelper(root.right)\n",
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" \n",
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" def printTree(self):\n",
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" self.printTreeHelper(self.root)\n",
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" \n",
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" \n",
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" def isPresentHelper(self,root,data):\n",
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" if root==None:\n",
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" return False\n",
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" if root.data==data:\n",
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" return True\n",
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" if root.data>data:\n",
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" #call on left\n",
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" return self.isPresentHelper(root.left,data)\n",
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" else:\n",
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" #Call on right\n",
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" return self.isPresentHelper(root.right,data)\n",
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" \n",
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" def search(self, data):\n",
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" return self.isPresentHelper(self.root,data)\n",
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" \n",
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" \n",
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" def insertHelper(self,root,data):\n",
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" if root==None:\n",
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" node=BinaryTreeNode(data)\n",
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" return node\n",
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" if root.data>=data:\n",
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" root.left=self.insertHelper(root.left,data)\n",
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" return root\n",
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" else:\n",
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" root.right=self.insertHelper(root.right,data)\n",
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" return root\n",
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" \n",
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" def insert(self, data):\n",
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" self.numNodes += 1\n",
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" self.root=self.insertHelper(self.root,data)\n",
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" \n",
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" \n",
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" def min(self,root):\n",
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" if root==None:\n",
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" return 10000\n",
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" if root.left==None:\n",
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" return root.data\n",
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" \n",
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" def deleteDataHelper(self,root,data):\n",
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" if root==None:\n",
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" return False, None\n",
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" if root.data<data:\n",
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" deleted,newRightNode=self.deleteDataHelper(root.right,data)\n",
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" root.right=newRightNode\n",
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" return deleted,root\n",
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" if root.data>data:\n",
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" deleted,newLeftNode=self.deleteDataHelper(root.left,data)\n",
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" root.left=newLeftNode\n",
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" return deleted,root\n",
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" #root is leaf\n",
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" if root.left==None and root.right==None:\n",
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" return True, None\n",
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" # root has one child\n",
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" if root.left==None:\n",
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" return True,root.right\n",
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" if root.right==None:\n",
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" return True,root.left\n",
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" #root has 2 children\n",
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" replacement=self.min(root.right)\n",
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" root.data=replacement\n",
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" deleted,newRightNode=self.deleteDataHelper(root.right,replacement)\n",
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" root.right=newRightNode\n",
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" return True,root\n",
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" \n",
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" def delete(self, data):\n",
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" deleted,newRoot=self.deleteDataHelper(self.root,data)\n",
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" if deleted:\n",
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" self.numNodes-=1\n",
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" self.root=newRoot\n",
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" return deleted\n",
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" \n",
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" def count(self):\n",
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" return self.numNodes\n",
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"\n",
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" \n",
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"b = BST()\n",
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"q = int(input())\n",
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"while (q > 0) :\n",
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" li = [int(ele) for ele in input().strip().split()]\n",
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" choice = li[0]\n",
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" q-=1\n",
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" if choice == 1:\n",
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" data = li[1]\n",
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" b.insert(data)\n",
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" elif choice == 2:\n",
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" data = li[1]\n",
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" b.delete(data)\n",
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" elif choice == 3:\n",
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" data = li[1]\n",
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" ans = b.search(data)\n",
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" if ans is True:\n",
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" print('true')\n",
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" else:\n",
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" print('false')\n",
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" else:\n",
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" b.printTree()"
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],
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"metadata": {
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"id": "uoTa5z4DzIJC"
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},
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"execution_count": null,
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"outputs": []
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}
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]
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}

‎07_Binary-Search-Tree/Binary-Search-Trees-2/Quiz_Binary-Search-Tree-2.pdf

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