题目描述

Invert a binary tree.

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

to

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

解法一

思路

这是一道经典题目了，自从Max Howell面试Google没有写出来这道题，就吸引了更高的人气了。

注意观察示例可以发现，简单来说反转二叉树就是对每个结点来说，调换左右子结点的位置，所以其实就是一个简单的二叉树遍历。那么二叉树的前序、中序和后序遍历里面该用哪种呢？可以看到每次操作（对于某个结点）只与“左”子结点和“右”子结点相关，即不影响“左右”的位置关系即可，所以这里用前序和后序遍历都可以。为什么不能用中序遍历？拿示例用中序遍历模拟一下就明白啦～

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def invertTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root is None:
            return
        self.invertTree(root.left)
        self.invertTree(root.right)
        root.left, root.right = root.right, root.left
        return root

Java

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

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *invertTree(TreeNode *root) {
        if (root == nullptr) {
            return nullptr;
        }
        invertTree(root->left);
        invertTree(root->right);
        TreeNode *temp = root->left;
        root->left = root->right;
        root->right = temp;
        return root;
    }
};