LeetCode 116.populating-next-right-pointers-in-eac

Posted

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了LeetCode 116.populating-next-right-pointers-in-eac相关的知识,希望对你有一定的参考价值。

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:
技术图片
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;

    }

}

以上是关于LeetCode 116.populating-next-right-pointers-in-eac的主要内容,如果未能解决你的问题,请参考以下文章

如何做LeetCode

leetcode可以写在简历上吗

[Leetcode]leetcode1-10题随记

leetcode分类刷题(续2)

leetcode分类刷题

LintCode,hihoCoder,LeetCode有啥区别