题目描述

给定一个二叉树

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。

示例：

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

思路

1.思路与116. 填充每个节点的下一个右侧节点指针类似，都是上一层遍历连接下一层。
2.由于从满二叉树变成了二叉树，所以需要作出以下几个改动

  • 设置每一层的头结点为亚结点，当亚结点的next为空时，结束遍历
  • cur在连接结点前，需要先判断该结点是否为null

Java代码实现

    public Node connect(Node root) {
        Node cur = root;
        Node p;
        while(cur != null){
            //每一层的头结点，是个亚结点
            p = new Node();
            Node head = p;
            while(cur != null){
                if(cur.left != null){
                    p.next = cur.left;
                    p = p.next;
                }
                if(cur.right != null){
                    p.next = cur.right;
                    p = p.next;
                }
                cur = cur.next;
            }
            //开始遍历下一层的结点
            cur = head.next;
        }
        return root;
    }