LeetCode 226. Invert Binary Tree

Problem

226. Invert Binary Tree

1. 题目简述

翻转一棵二叉树。例如：

Invert a binary tree.

Example:

Input:

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

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

2. 算法思路

一道被大佬吐槽的easy题目。DFS或者BFS均可。

DFS

递归调用invert，前中后序均可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    // 前序
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }

        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        if (root.left != null) {
            root.left = invertTree(root.left);
        }
        if (root.right != null) {
            root.right = invertTree(root.right);
        }

        return root;
    }

    // 中序
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }

        if (root.left != null) {
            root.left = invertTree(root.left);
        }

        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        if (root.right != null) {
            root.right = invertTree(root.right);
        }

        return root;
    }
    //后序
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }

        if (root.left != null) {
            root.left = invertTree(root.left);
        }
        if (root.right != null) {
            root.right = invertTree(root.right);
        }

        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;

        return root;
    }
}

BFS

class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return root;
        }

        Queue<TreeNode> queue = new LinkedList();
        queue.add(root);

        while (!queue.isEmpty()) {
            TreeNode node = queue.poll();
            TreeNode temp = node.left;
            node.left = node.right;
            node.right = temp;

            if (node.left != null) {
                queue.add(node.left);
            }
            if (node.right != null) {
                queue.add(node.right);
            }
        }

        return root;
    }
}