[Leetcode] Populating Next Right Pointers in Each Node ii

liuwenyirachel

于 2013-03-02 21:32:11 发布

阅读量166

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/liuwenyirachel/article/details/8630283

Leetcode 专栏收录该内容

67 篇文章

订阅专栏

本文探讨了如何在任意二叉树中填充每个节点的Next指针，使其指向同层的下一个节点。解决方案需确保仅使用常数额外空间，并通过递归方式实现。文章还提供了一个C++示例代码，演示了如何完成这一任务。

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

用递归，这样算是 Constant Space 吗？

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root==NULL) return;
        
        TreeLinkNode* p = root;
        TreeLinkNode* q = NULL;
        TreeLinkNode* nextLevel = NULL;
        
        while(p)
        {
            if(p->left)
            {
                if(q)
                {
                    q->next = p->left;
                }
                q = p->left;
                if(nextLevel==NULL)
                {
                    nextLevel = p->left;
                }
            }
            if(p->right)
            {
                if(q)
                {
                    q->next = p->right;
                }
                q = p->right;
                if(nextLevel==NULL)
                {
                    nextLevel = p->right;
                }
            }
            p = p->next;
        }   
        connect(nextLevel);
    }
};