LeetCode：Populating Next Right Pointers in Each Node

最新推荐文章于 2018-01-23 21:53:49 发布

hackerzchao

最新推荐文章于 2018-01-23 21:53:49 发布

阅读量489

点赞数

CC 4.0 BY-SA版权

分类专栏： LeetCode LeetCode 文章标签： algorithm leetcode

本文链接：https://blog.youkuaiyun.com/hnuzengchao/article/details/42030131

LeetCode 同时被 2 个专栏收录

48 篇文章

订阅专栏

LeetCode

48 篇文章

订阅专栏

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。通过递归方式实现，确保只使用常数额外空间。

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

递归一定还要学会好好运用。

// Source : https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
// Author : Chao Zeng  
// Date   : 2014-12-19 
struct TreeLinkNode {
	int val;
	TreeLinkNode *left, *right, *next;
	TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

class Solution {
public:
	void connect(TreeLinkNode *root) {
		if (!root)
			return;
		if (root->left != NULL)
			root->left->next = root->right;
		if (root->right != NULL)
			root->right->next = root->next ? root->next->left : NULL;
		connect(root->left);
		connect(root->right);
	}
};