说明

算法：Invert Binary Tree
LeetCode地址：https://leetcode.com/problems/invert-binary-tree/

题目：
Invert a binary tree.

Example:

Input:

     4/   \2     7/ \   / \
1   3 6   9

Output:

     4/   \7     2/ \   / \
9   6 3   1

Trivia:
This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

解题思路1

翻转二叉树，深度优先遍历，用递归的方式翻转，注意到层级一致来调换，所以先把递归调用写在前面，最后再用交换。
写二叉树比较容易，校验代码如何写。笔者按照例子入参，改造为TreeNode对象，再改写TreeNode的 ToString() 方法，把输入和输出的二叉树都已数组的方式打印出来。画二叉树的规则是，逐层从上到下画出即可。
时间复杂度为 O(N), 数据都会遍历一遍，时间复杂度跟层数成正比，复杂度在log(n) ~ n 之间。

代码实现1

import java.util.LinkedList;
import java.util.Queue;class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x; }@Overridepublic String toString() {Queue<TreeNode> queue = new LinkedList<>();queue.add(this);StringBuilder sb = new StringBuilder();sb.append("[");while (!queue.isEmpty()) {TreeNode current = queue.poll();sb.append(current.val).append(", ");if (current.left != null) queue.add(current.left);if (current.right != null) queue.add(current.right);}sb.append("]");return sb.toString();}
}public class InvertBinaryTree {public TreeNode invertTree(TreeNode root) {if (root == null) {return root;}TreeNode left = invertTree(root.left);TreeNode right = invertTree(root.right);root.left = right;root.right = left;return root;}public static TreeNode initData() {int[] data = {4, 2, 7, 1, 3, 6, 9};TreeNode root = new TreeNode(4);Queue<TreeNode> queue = new LinkedList<>();queue.add(root);for (int i = 1; i < data.length; ) {TreeNode currentRoot = queue.poll();TreeNode left = new TreeNode(data[i]);currentRoot.left = left;queue.add(currentRoot.left);if (i + 1 < data.length) {TreeNode right = new TreeNode(data[++i]);currentRoot.right = right;queue.add(currentRoot.right);}i++;}return root;}public static void main(String[] args) {TreeNode root = initData();System.out.println("Input: " + root.toString());InvertBinaryTree obj = new InvertBinaryTree();obj.invertTree(root);System.out.println("Output: " + root.toString());}}

运行结果1

Input: [4, 2, 7, 1, 3, 6, 9, ]
Output: [4, 7, 2, 9, 6, 3, 1, ]

代码执行效率1

Runtime: 0 ms, faster than 100.00% of Java online submissions for Invert Binary Tree.
Memory Usage: 35.7 MB, less than 29.71% of Java online submissions for Invert Binary Tree.

解题思路2

用递归的方式实现的风险是，方法指针栈会导致栈溢出，所以改为内存栈的方式，按照广度优先遍历的方式处理。
时间复杂度为 O(N)。

代码实现2

import java.util.LinkedList;
import java.util.Queue;public class InvertBinaryTree {public TreeNode invertTreeWithBreadthSearchFirst(TreeNode root) {if (root == null) {return root;}Queue<TreeNode> queue = new LinkedList<>();queue.add(root);while (!queue.isEmpty()) {TreeNode current = queue.poll();TreeNode temp =current.left;current.left = current.right;current.right = temp;if (current.left != null) queue.add(current.left);if (current.right != null) queue.add(current.right);}return root;}}

运行结果2

Input: [4, 2, 7, 1, 3, 6, 9, ]
Output: [4, 7, 2, 9, 6, 3, 1, ]

代码执行效率2

Runtime: 0 ms, faster than 100.00% of Java online submissions for Invert Binary Tree.
Memory Usage: 35.8 MB, less than 6.82% of Java online submissions for Invert Binary Tree.

总结

翻转二叉树有两种方式，一种是深度优先遍历，第二种是广度优先遍历。

代码下载：
https://github.com/zgpeace/awesome-java-leetcode/blob/master/code/LeetCode/src/popular/InvertBinaryTree.java

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算法:Invert Binary Tree(翻转二叉树)

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