Leetcode 156. Binary Tree Upside Down (Medium) (cpp)

最新推荐文章于 2022-06-06 00:04:15 发布

原创最新推荐文章于 2022-06-06 00:04:15 发布 · 536 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #cpp

C++ 同时被 3 个专栏收录

299 篇文章

订阅专栏

Leetcode C++

287 篇文章

订阅专栏

Leetcode Tree

48 篇文章

订阅专栏

本文介绍了一种特殊的二叉树翻转算法，即将特定形态的二叉树翻转为新的形态，使得原来的右子节点变为左子叶节点。通过递归方式实现，最终返回新的根节点。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Leetcode 156. Binary Tree Upside Down (Medium) (cpp)

Tag: Tree

Difficulty: Medium

/*

156. Binary Tree Upside Down (Medium)

Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) or empty, flip it upside down and turn it into a tree where the original right nodes turned into left leaf nodes. Return the new root.

For example:
Given a binary tree {1,2,3,4,5},
1
/ \
2   3
/ \
4   5
return the root of the binary tree [4,5,2,#,#,3,1].
4
/ \
5   2
/ \
3   1

*/
/**
* 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* upsideDownBinaryTree(TreeNode* root) {
		if (root == NULL || root->left == NULL) {
			return root;
		}
		TreeNode *node = upsideDownBinaryTree(root->left);
		root->left->left = root->right;
		root->left->right = root;
		root->left = NULL;
		root->right = NULL;
		return node;
	}
};