Invert Binary Tree

https://leetcode.com/problems/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.

解题思路：

这道题来源于上面写的一条tweet，其实是很简单的题目。Symmetric Tree 这道题里有很类似的思路，下面写一个递归的方法。这道题的返回值是TreeNode，当然也可以用void的返回值。

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

若不用递归，就更简单了，虽然代码复杂点。但是会BFS的应该都会，一模一样的便利，只不过多了一个将每个节点左右子节点交换的操作。

/**
 * 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 root;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while(queue.size() > 0) {
            TreeNode cur = queue.poll();
            TreeNode temp = cur.left;
            cur.left = cur.right;
            cur.right = temp;
            if(cur.left != null) {
                queue.offer(cur.left);
            }
            if(cur.right != null) {
                queue.offer(cur.right);
            }
        }
        return root;
    }
}