LeetCode-Go/leetcode/0116.Populating-Next-Right-Pointers-in-Each-Node

116. Distinct Subsequences

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.

Follow up:

• 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.

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.

Constraints:

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

题目大意

将二叉树的每一层节点都连接起来形成一个链表，每层的最后一个节点指向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

package leetcode

// 解法一：迭代
func connect(root *Node) *Node {
	if root == nil {
		return root
	}

	q := []*Node{root}

	for len(q) > 0 {
		var p []*Node

		for i, node := range q {
			if i+1 < len(q) {
				node.Next = q[i+1]
			}
			if node.Left != nil {
				p = append(p, node.Left)
			}
			if node.Right != nil {
				p = append(p, node.Right)
			}
		}
		q = p
	}

	return root
}


// 解法二 递归
func connect2(root *Node) *Node {
	if root == nil {
		return nil
	}
	connectTwoNode(root.Left, root.Right)

	return root
}

func connectTwoNode(node1, node2 *Node) {
	if node1 == nil || node2 == nil {
		return
	}
	node1.Next = node2

	connectTwoNode(node1.Left, node1.Right)
	connectTwoNode(node2.Left, node2.Right)
	connectTwoNode(node1.Right, node2.Left)
}