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。

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":{"$id":"5","left":{"$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}

思路

1.第一层连接第二层，第二层连接第三层，依次类推，迭代到最后。
2.使用cur标识本层正在移动的结点，pre标识cur结点的前一个结点，start标识本层第一个结点。
3.具体迭代过程，可以参考下图

第一次

第二次

第三次

Java代码实现

public Node connect(Node root) {
        if(root == null)
            return null;

        //当前遍历的前一个节点
        Node pre = root;
        //当前遍历的节点
        Node cur = null;
        //每层开始的节点
        Node start = pre;

        while(start.left != null){
            //如果本层走到了NULL，就代表需要进行下一层的处理
            if(cur == null){
                //连接上最后的两个节点
                pre.left.next = pre.right;
                //初始化pre节点
                pre = start.left;
                //初始化cur节点
                cur = start.right;
                //初始化start节点
                start = pre;
           }
            //依次连接下层节点
            else{
                pre.left.next = pre.right;
                pre.right.next = cur.left;
                pre = pre.next;
                cur = cur.next;
            }
        }
        return root;
    }