Merge pull request #5 from born69confused/patch-1

1367.Linked List in Binary Tree

Author: Jack-Cherish

Linked List/Medium/1367.Linked List in Binary Tree

@@ -0,0 +1,73 @@
+QUESTION :
+Given a binary tree root and a linked list with head as the first node. 
+Return True if all the elements in the linked list starting from the head correspond to some downward path connected in the binary tree otherwise return False.
+In this context downward path means a path that starts at some node and goes downwards.
+
+Example 1:
+Input: head = [4,2,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: true
+Explanation: Nodes in blue form a subpath in the binary Tree.  
+
+Example 2:
+Input: head = [1,4,2,6], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: true
+
+Example 3:
+Input: head = [1,4,2,6,8], root = [1,4,4,null,2,2,null,1,null,6,8,null,null,null,null,1,3]
+Output: false
+Explanation: There is no path in the binary tree that contains all the elements of the linked list from head.
+ 
+Constraints:
+1 <= node.val <= 100 for each node in the linked list and binary tree.
+The given linked list will contain between 1 and 100 nodes.
+The given binary tree will contain between 1 and 2500 nodes.
+
+C++ SOLUTION :
+
+class Solution {
+public:
+    bool isSubPath(ListNode* head, TreeNode* root) {
+        if(root){                                                            
+            if(isEqual(head, root)){
+                return true;
+            }
+            return isSubPath(head, root->left) || isSubPath(head, root->right);
+        }
+        return false;
+    }
+    bool isEqual(ListNode* head, TreeNode* root){
+        if(!head){
+            return true;
+        }
+        if(!root){
+            return false;
+        }
+        return head->val == root->val && (isEqual(head->next, root->left) || isEqual(head->next, root->right));
+    }
+};
+
+PYTHON SOLUTION :
+
+def isSubPath(self, head: ListNode, root: TreeNode) -> bool:
+    target = ""
+    while head:
+        target = target + str(head.val)
+        head = head.next
+    
+    def dfs(root, path):
+        if target in path:
+            return True
+            
+        if root.left:
+            ans =  dfs(root.left, path + str(root.left.val))
+            if ans == True:
+                return True
+                
+        if root.right:
+            ans  = dfs(root.right, path + str(root.right.val))
+            if ans == True:
+                return True
+        
+        return False
+        
+    return dfs(root, str(root.val))