leetcode：Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

原创最新推荐文章于 2021-09-24 12:05:40 发布 · 460 阅读

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本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使其指向右邻居节点。文章提供了一个高效的解决方案，仅使用常数额外空间，并假设输入为完美二叉树。

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题目描述：

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *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.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

不得不说：刚开始先是想到了用队列层次遍历，后来发现还是有更优先的代码；

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
           TreeLinkNode level_start=root;
        while(level_start!=null){
            TreeLinkNode cur=level_start;
            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;
            }
            level_start=level_start.left;
        }
    }
}