题目描述

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。

样例

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

提示：

• 你只能使用常量级额外空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

算法分析

bfs

• 1、用队列将当前层的结点全部存起来，并记录有len个是属于当前层
• 2、将当前层的每个结点都指向下一个结点，poll出该结点，并把结点的左右儿子也加入到队列中，为下一层做准备
• 3、当前层的最后一个结点需要指向空

时间复杂度 $O(n)$

Java 代码

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if(root == null) return root;
        Queue<Node> q = new LinkedList<Node>();
        q.add(root);
        while(!q.isEmpty())
        {
            int len = q.size();
            for(int i = 0;i < len - 1;i ++)
            {
                Node t = q.poll();
                t.next = q.peek();
                if(t.left != null) q.add(t.left);
                if(t.right != null) q.add(t.right);
            }
            //将最后一个元素指向null
            Node t = q.poll();
            t.next = null;
            if(t.left != null) q.add(t.left);
            if(t.right != null) q.add(t.right);
        }
        return root;
    }
}