0%

### 题目描述

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

`struct Node {`
`int val;`
`Node *left;`
`Node *right;`
`Node *next;`
`}`
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

### 例子

#### 例子 1

Input: `root = [1,2,3,4,5,6,7]`
Output: `[1,#,2,3,#,4,5,6,7,#]`
Explanation:
Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with ‘#’ signifying the end of each level.

You may only use constant extra space.
Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

### Constraints

The number of nodes in the given tree is less than `4096`.
`-1000 <= node.val <= 1000`

### 解题思路

• 对于根节点 `root`, 调用 `connect(root->left, root->right)`
• `connect` 函数中：
• 首先连接输入的两个节点： `left->next = right`
• 然后执行以下操作即可：
• `connect(left->left, left->right)`
• `connect(left->right, right->left)`
• `connect(right->left, right->right)`

• 时间复杂度: O(n)
• 空间复杂度: O(h) — 递归深度

GitHub 代码同步地址： 116.PopulatingNextRightPointersInEachNode.cpp

GitHub: Leetcode-C++-Solution