(easy)LeetCode 226.Invert Binary Tree

標籤：

Invert a binary tree.

     4   /     2     7 / \   / 1   3 6   9

to

     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 fuck off.

代碼如下：

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public  TreeNode invertTree(TreeNode root) {if (root == null || (root.left == null) && (root.right == null))return root;root=invertTreeLR(root,root.left, root.right);return root;}public  TreeNode invertTreeLR(TreeNode root,TreeNode left, TreeNode right) { //left,right位臨時引用if (left == null && right != null) {left = right;right = null;left=invertTreeLR(left,left.left, left.right);} else if (left != null && right == null) {right = left;left = null;right=invertTreeLR(right,right.left, right.right);} else if (left == null && right == null){}else {TreeNode tmp = null;tmp = left;left = right;right = tmp;left=invertTreeLR(left,left.left, left.right);right=invertTreeLR(right,right.left, right.right);}root.left=left;root.right=right;return root;}}

　　運行結果：

  

(easy)LeetCode 226.Invert Binary Tree