226 翻转二叉树 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.

翻转二叉树
  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     public int val;
  5.  *     public TreeNode left;
  6.  *     public TreeNode right;
  7.  *     public TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  10. public class Solution {
  11.     public TreeNode InvertTree(TreeNode root) {
  12.         if(root==null||(root.left==null&&root.right==null)) return root;
  13.         var tem = root.left;
  14.         root.left = root.right;
  15.         root.right = tem;
  16.         if(root.right!=null){
  17.             InvertTree(root.right);
  18.         }
  19.         if(root.left!=null){
  20.             InvertTree(root.left);
  21.         }
  22.         return root;
  23.     }
  24. }

payhon BFS版
  1. class Solution(object):
  2.     def invertTree(self, root):
  3.         """
  4.         :type root: TreeNode
  5.         :rtype: TreeNode
  6.         """
  7.         if (not root):
  8.             return
  9.         queue = [root]
  10.         while (queue):
  11.             node = queue.pop(0)
  12.             node.left,node.right = node.right,node.left
  13.             if(node.left):
  14.                 queue.append(node.left)
  15.             if(node.right):
  16.                 queue.append(node.right)
  17.         return root







posted @ 2017-01-10 22:58  xiejunzhao  阅读(136)  评论(0)    收藏  举报