LeetCode 116.populating-next-right-pointers-in-each-node

Posted 2021-03-09 swineherd_MCQ

题意

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。

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}

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。

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路

题目要求使用O（1）的额外空间，所以考虑类似BFS的算法。

因为树是完美的，那么当前这一层和上一层的关系是紧密的，体现在上一层节点cur存在next不为null那么当前层cur.left也存在next并且cur.right也存在next，可以根据示例图理解。每一层从上一层的最左边节点的左孩子开始遍历。

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) {

        Node pre=root;

        while(pre!=null){

            Node cur=pre;

            while(cur!=null){

                if(cur.left!=null)

                    cur.left.next=cur.right;

                if(cur.right!=null&&cur.next!=null){

                    cur.right.next=cur.next.left;

                }

                cur=cur.next;

            }

            pre=pre.left;

        }

        return root;

    }

}