116. Populating Next Right Pointers in Each Node（js）

完美二叉树节点连接算法

最新推荐文章于 2022-02-28 18:16:11 发布

转载最新推荐文章于 2022-02-28 18:16:11 发布 · 48 阅读

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原文链接：http://www.cnblogs.com/xingguozhiming/p/10777706.html

本文介绍了一种算法，用于在完美二叉树中填充每个节点的next指针，使其指向其右侧的下一个节点。如果右侧没有节点，则next指针设置为NULL。通过递归方法实现，首先设置根节点的next指针，然后分别处理左子树和右子树。

116. Populating Next Right Pointers in Each Node

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.
题意：将每个树节点向右连接，如果右边无节点则指向空
代码如下：

/**
 * Definition for binary tree with next pointer.
 * function TreeLinkNode(val) {
 *     this.val = val;
 *     this.left = this.right = this.next = null;
 * }
 */

/**
 * @param {TreeLinkNode} root
 * @return {void} Do not return anything, modify tree in-place instead.
 */
var connect = function(root) {
    return connectChild(root,null)
};
var connectChild=function(root,next){
    if (!root) return root;
    root.next = next;
    connectChild(root.left, root.right);
    connectChild(root.right, (root.next ? root.next.left : root.next));
    return root;
}

转载于:https://www.cnblogs.com/xingguozhiming/p/10777706.html