Invert Binary Tree

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.

递归后序遍历, 因为有返回值, 所以要后序遍历, 在递归回溯后返的时候进行操作:

/**
 * 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) {
            return null;
        }
        TreeNode left = root.left;
        TreeNode right = root.right;
       
        root.left = invertTree(right);
        root.right = invertTree(left);
        return root;
    }

一般在改变递归函数的输入值的时候加上 这句, 防止递归两次null改变两次输入值

if (root.left == null && root.right == null) {
return root;
}

先序遍历

public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            TreeNode left = node.left;
            TreeNode right = node.right;
            if (left != null) {
                stack.push(left);
            }
            if (node.right != null) {
                stack.push(node.right);
            }
            node.left = right;
            node.right = left;
           
        }
        return root;
    }