LeetCode oj 226. Invert Binary Tree （DFS||BFS）

226. Invert Binary Tree

 

 My Submissions

• Total Accepted: 124813
• Total Submissions: 257083
• Difficulty: Easy

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.

二叉樹的翻轉

做了幾道關於這方面的題，感覺自己已經成爲了一名合格的司機，例行提供DFS或BFS的兩種題解

DFS：

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

BFS：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public TreeNode invertTree(TreeNode root) {
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        if(root == null)
            return null;
        q.clear();
        q.add(root);
        while(!q.isEmpty()){
            TreeNode temp_root = q.poll();
            TreeNode temp_left = temp_root.left;
            temp_root.left = temp_root.right;
            temp_root.right = temp_left;
            if(temp_root.left != null) 
                q.add(temp_root.left);
            if(temp_root.right != null) 
                q.add(temp_root.right); 
        }
        return root;
    }
}