题目描述:
操作给定的二叉树,将其变换为源二叉树的镜像。
如下图所示:
解题思路:
先交换根节点的两个子结点之后,我们注意到值为10、6的结点的子结点仍然保持不变,因此我们还需要交换这两个结点的左右子结点。做完这两次交换之后,我们已经遍历完所有的非叶结点。此时变换之后的树刚好就是原始树的镜像。交换示意图如下所示:
举例:
代码实现(c++)
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};*/
class Solution {
public:
void Mirror(TreeNode *pRoot) {
if((pRoot == NULL) || (pRoot->left == NULL && pRoot->right == NULL)){
return;
}
//交换根节点的左右结点
TreeNode *pTemp = pRoot->left;
pRoot->left = pRoot->right;
pRoot->right = pTemp;
//递归左子树
if(pRoot->left){
Mirror(pRoot->left);
}
//递归右子树
if(pRoot->right){
Mirror(pRoot->right);
}
}
};
代码实现(java)
/**
public class TreeNode {
int val = 0;
TreeNode left = null;
TreeNode right = null;
public TreeNode(int val) {
this.val = val;
}
}
*/
public class Solution {
public void Mirror(TreeNode root) {
//当前节点为空,直接返回
if(root == null)
return;
//当前节点没有叶子节点,直接返回
if(root.left == null && root.right == null)
return;
TreeNode temp = root.left;
root.left = root.right;
root.right = temp;
//递归交换叶子节点
if(root.left != null)
Mirror(root.left);
if(root.right != null)
Mirror(root.right);
}
}
代码实现(Python2.7)
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回镜像树的根节点
def Mirror(self, root):
# write code here
if (root == None or (root.left == None and root.right == None)):
return None
tmp = root.left
root.left = root.right
root.right = tmp
if root.left:
self.Mirror(root.left)
if root.right:
self.Mirror(root.right)