[LeetCode] Populating Next Right Pointers in Each Node

最新推荐文章于 2025-12-06 07:43:37 发布

原创最新推荐文章于 2025-12-06 07:43:37 发布 · 264 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm #leetcode #算法

Algorithm 同时被 2 个专栏收录

12 篇文章

订阅专栏

LeetCode

9 篇文章

订阅专栏

本文详细阐述了如何使用递归方法实现二叉树中节点的连接，包括左子节点与右子节点、父节点之间的连接关系，适用于完美二叉树场景。

题目：

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

题解：

如果不限定constant extra space. 思路很简单，采用层次遍历的方法就可以解决，设一个队列，父节点出，子节点进，一步一步链接即可，时间复杂度和空间复杂度都是O(n)

题目限制了额外空间，很直接的想法就是从上往下一层一层的加指针，左孩子的next指针就是指向父节点的右孩子，右孩子的next节点就是指向父节点的next节点的左孩子。这样，一步一步递归下去，就可以得到了。

/**
 * Definition for binary tree with next pointer.
 * 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 == NULL)
			return;
		if(root->left)
			root->left->next = root->right;
		if(root->right)
			root->right->next =(root->next)? root->next->left:NULL;
		connect(root->left);
		connect(root->right);
	}
};

但我之前的代码啥时如下写的，提示runtime error，last executed input: {}。引以为戒！

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