LeetCode 116. Populating Next Right Pointers in Each Node

最新推荐文章于 2023-11-09 13:02:00 发布

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最新推荐文章于 2023-11-09 13:02:00 发布

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本文链接：https://blog.youkuaiyun.com/qq_39638957/article/details/91384778

本文介绍了一种算法，用于填充完美二叉树中每个节点的next指针，使其指向同一层级的下一个节点。通过迭代和链接的方式，算法在常数额外空间下实现了这一目标，适用于那些所有叶子节点都在同一水平且每个父节点都有两个子节点的二叉树结构。

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You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example:

Input: {"$id":"1","left":{"$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}

Output: {"$id":"1","left":{"$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}

Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B.

Note:

You may only use constant extra space.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.

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

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    
    /*
    root是从根节点出发，一路向左节点走
    cur则是在每一层，一路从左往右遍历该层节点
    */
    
    public Node connect(Node root) {
        
        if(root==null) return null;
        
        Node res = root;//copy root
        
        while(root.left != null){//第0层,直到倒数第二层
            Node cur = root;
            while(cur != null) {//对该层的每一个node进行处理
                cur.left.next = cur.right;
                cur.right.next = cur.next == null ? null : cur.next.left;
                cur = cur.next;//向右走
            }
            root = root.left;
        }
        
        return res;
    }
    
}