116. 填充每个节点的下一个右侧节点指针

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题目

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.

Example 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.

Example 2:

Input: root = []

Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 212 - 1].
• -1000 <= Node.val <= 1000

Follow-up:

• You may only use constant extra space.
• The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

题目大意

给定一个 完美二叉树 ，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。初始状态下，所有  next 指针都被设置为 NULL。

解题思路

本质上是二叉树的层序遍历，基于广度优先搜索，将每层的节点放入队列，并遍历队列进行连接。

代码

/**
 * @param {Node} root
 * @return {Node}
 */
var connect = function (root) {
	if (!root) return root;
	let queue = [root];
	while (queue.length) {
		let len = queue.length;
		for (let i = 0; i < len; i++) {
			if (i + 1 < len) queue[i].next = queue[i + 1];
			if (queue[i].left) queue.push(queue[i].left);
			if (queue[i].right) queue.push(queue[i].right);
		}
		queue = queue.slice(len);
	}
	return root;
};

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