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atwood
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3 files changed

+104
-39
lines changed

3 files changed

+104
-39
lines changed

‎INo64_Math.cs

+5
Original file line numberDiff line numberDiff line change
@@ -20,6 +20,11 @@ namespace LeetCode_I64
2020
// }
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//}
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/// <summary>
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/// REVIEW
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/// 2020.08.03: 要求的难点就是不能使用if等关键字，但是其实也是用了，只不过是用了另外的方法躲开了检测而已。
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/// </summary>
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/// <summary>
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/// Unsolved Question: 为什么使用了类中全局的变量，避免了递归每次申请变量，反而内存消耗变大了。这个递归应该会申请很多个bool才对啊
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/// </summary>

‎No114_Tree.cs

+96
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,96 @@
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using LeetCode.ExtensionFunction;
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using System;
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using System.Collections.Generic;
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using System.Text;
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namespace LeetCode_114
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{
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//static void Main(string[] args)
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//{
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// var solution = new Solution();
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// while (true)
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// {
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// //int input = int.Parse(Console.ReadLine());
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// //int input2 = int.Parse(Console.ReadLine());
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// //string input = Console.ReadLine();
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// //string input2 = Console.ReadLine();
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// //int[] intArr = input.Split(',').Select(s => int.Parse(s)).ToArray();
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// //int input2 = int.Parse(Console.ReadLine());
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// var builder = new DataStructureBuilder();
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// int?[] data = new int?[] { 1, 2, 5, 3, 4, null, 6 };
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// var tree = builder.BuildTree(data);
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// //var listNode = builder.BuildListNode(new int[] { 1, 4, 5 });
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// //int[][] arr = new int[3][] { new int[] { 1, 3, 1 }, new int[] { 1, 5, 1 }, new int[] { 4, 2, 1 } };
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// //int[] nums1 = new int[] { 2, 1, -2, 3 };
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// //int[] nums2 = new int[] { 10, 15, 20 };
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// //string input = "adceb";
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// //string input2 = "*a*b";
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// //var res = solution.IsMatch(input, input2);
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// //ConsoleX.WriteLine(res);
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// solution.Flatten(tree);
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// }
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//}
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public class Solution
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{
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/// <summary>
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/// 运用了迭代中序遍历的代码，先把节点都入队，然后串联成目标树就可以了
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/// 时间复杂度：O(n)
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/// 空间复杂度：O(n)
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/// </summary>
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/// <param name="root"></param>
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public void Flatten(TreeNode root)
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{
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Queue<TreeNode> queue = new Queue<TreeNode>();
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Stack<TreeNode> stack = new Stack<TreeNode>();
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TreeNode curr = root;
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while (curr != null || stack.Count > 0)
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{
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while (curr != null)
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{
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queue.Enqueue(curr);
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stack.Push(curr);
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curr = curr.left;
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}
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curr = stack.Pop();
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curr = curr.right;
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}
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if (queue.Count != 0)
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root = queue.Dequeue();
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while (queue.Count != 0)
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{
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//这里有一点要注意的，一定要用root.right去串联，如果用root去串联的话，其实只是变量指向的空间而已，并没有串起来
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root.right = queue.Dequeue();
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root.left = null;
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root = root.right;
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}
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}
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/// <summary>
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/// 类似莫里斯遍历的方法,通过直接改变树结构的方式完成
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/// 时间复杂度：O(n)
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/// 空间复杂度：O(1)
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/// </summary>
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/// <param name="root"></param>
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//public void Flatten(TreeNode root)
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//{
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// TreeNode curr = root;
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// while (curr != null)
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// {
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// if (curr.left != null)
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// {
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// TreeNode next = curr.left;
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// TreeNode predecessor = next;
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// while (predecessor.right != null)
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// {
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// predecessor = predecessor.right;
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// }
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// predecessor.right = curr.right;
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// curr.left = null;
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// curr.right = next;
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// }
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// curr = curr.right;
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// }
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//}
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}
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}

‎Program.cs

+3-39
Original file line numberDiff line numberDiff line change
@@ -21,9 +21,9 @@ static void Main(string[] args)
2121
//string input2 = Console.ReadLine();
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//int[] intArr = input.Split(',').Select(s => int.Parse(s)).ToArray();
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//int input2 = int.Parse(Console.ReadLine());
24-
var builder = new DataStructureBuilder();
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int?[] data = new int?[] { 1, 2, 5, 3, 4, null, 6 };
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var tree = builder.BuildTree(data);
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//var builder = new DataStructureBuilder();
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//int?[] data = new int?[] { 1, 2, 5, 3, 4, null, 6 };
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//var tree = builder.BuildTree(data);
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//var listNode = builder.BuildListNode(new int[] { 1, 4, 5 });
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//int[][] arr = new int[3][] { new int[] { 1, 3, 1 }, new int[] { 1, 5, 1 }, new int[] { 4, 2, 1 } };
2929
//int[] nums1 = new int[] { 2, 1, -2, 3 };
@@ -32,45 +32,9 @@ static void Main(string[] args)
3232
//string input2 = "*a*b";
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//var res = solution.IsMatch(input, input2);
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//ConsoleX.WriteLine(res);
35-
solution.Flatten(tree);
3635
}
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}
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39-
public class Solution
40-
{
41-
/// <summary>
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/// 运用了迭代中序遍历的代码，先把节点都入队，然后串联成目标树就可以了
43-
/// 时间复杂度：O(n)
44-
/// 空间复杂度：O(n)
45-
/// </summary>
46-
/// <param name="root"></param>
47-
public void Flatten(TreeNode root)
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{
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Queue<TreeNode> queue = new Queue<TreeNode>();
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Stack<TreeNode> stack = new Stack<TreeNode>();
51-
TreeNode curr = root;
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while (curr != null || stack.Count > 0)
53-
{
54-
while (curr != null)
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{
56-
queue.Enqueue(curr);
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stack.Push(curr);
58-
curr = curr.left;
59-
}
60-
curr = stack.Pop();
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curr = curr.right;
62-
}
63-
if (queue.Count != 0)
64-
root = queue.Dequeue();
65-
while (queue.Count != 0)
66-
{
67-
//这里有一点要注意的，一定要用root.right去串联，如果用root去串联的话，其实只是变量指向的空间而已，并没有串起来
68-
root.right = queue.Dequeue();
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root.left = null;
70-
root = root.right;
71-
}
72-
}
73-
}
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//public class Solution
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//{
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// public bool IsMatch(string s, string p)

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