一、问题
https://leetcode-cn.com/problems/invert-binary-tree/
翻转一棵二叉树。 示例: 输入: 4 / 2 7 / \ / 1 3 6 9 输出: 4 / 7 2 / \ / 9 6 3 1 备注: 这个问题是受到 Max Howell 的 原问题 启发的 : 谷歌:我们90%的工程师使用您编写的软件(Homebrew),但是您却无法在面试时在白板上写出翻转二叉树这道题,这太糟糕了。
二、GitHub实现:https://github.com/JonathanZxxxx/LeetCode/blob/master/InvertTreeClass.cs
Blog:https://www.cnblogs.com/zxxxx/
三、思路
1、递归:互换左右孩子,对左右孩子进行递归
2、迭代:根节点入栈,每次迭代中,移除栈顶元素并互换左右孩子,如果当前节点有左孩子,入栈,为空不做处理,右孩子同理
四、代码实现
1 public class InvertTreeClass 2 { 3 /// <summary> 4 /// 递归 5 /// </summary> 6 /// <param name="root"></param> 7 /// <returns></returns> 8 public TreeNode InvertTree(TreeNode root) 9 { 10 if (root == null) 11 { 12 return null; 13 } 14 var left = InvertTree(root.right); 15 var right = InvertTree(root.left); 16 root.left = left; 17 root.right = right; 18 return root; 19 } 20 21 /// <summary> 22 /// 迭代 23 /// </summary> 24 /// <param name="root"></param> 25 /// <returns></returns> 26 public TreeNode InvertTree2(TreeNode root) 27 { 28 if (root == null) return null; 29 var stack = new Stack<TreeNode>(); 30 stack.Push(root); 31 while (stack.Any()) 32 { 33 var pop = stack.Pop(); 34 var temp = pop.left; 35 pop.left = pop.right; 36 pop.right = temp; 37 if (pop.left != null) stack.Push(pop.left); 38 if (pop.right != null) stack.Push(pop.right); 39 } 40 return root; 41 } 42 } 43 44 public class TreeNode 45 { 46 public int val; 47 public TreeNode left; 48 public TreeNode right; 49 public TreeNode(int x) { val = x; } 50 }