Populating Next Right Pointers in Each Node II

最新推荐文章于 2024-08-25 16:11:27 发布

Kehl

最新推荐文章于 2024-08-25 16:11:27 发布

阅读量612

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/Oliverkehl/article/details/39481593

leetcode 专栏收录该内容

65 篇文章

订阅专栏

针对Populating Next Right Pointers in Each Node问题的进阶版本，本文介绍了一种使用常数额外空间的方法来处理任意二叉树的情况，并提供了一个简洁的C++实现方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

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

使用一个虚拟头结点来解决问题，代码量小了很多

class Solution {
public:
    void connect(TreeLinkNode *root) {
        dfs(root);
    }
    void dfs(TreeLinkNode* node){
        if(node==NULL)
            return;
        TreeLinkNode* head = new TreeLinkNode(-1);
        TreeLinkNode* tail = head;
        while(node!=NULL){
            if(node->left!=NULL){
                tail->next = node->left;
                tail = tail->next;
            }
            if(node->right!=NULL){
                tail->next = node->right;
                tail = tail->next;
            }
            node= node->next;
        }
        dfs(head->next);
    }
};