[LeetCode] Populating Next Right Pointers in Each Node 解题报告

最新推荐文章于 2023-03-11 16:53:38 发布

原创最新推荐文章于 2023-03-11 16:53:38 发布 · 212 阅读

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本文介绍了一种仅使用常数额外空间填充二叉树中每个节点的next指针的方法，使其指向同一层的下一个节点。如果不存在下一个节点，则设置为NULL。此算法适用于完美二叉树，并通过递归方式实现。

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

» Solve this problem

[解题报告]
当前层处理完next指针的连接以后，再调用下一级节点。

[Code]

1:   void connect(TreeLinkNode *root) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      if(root == NULL) return;  
5:      if(root->left != NULL)  
6:        root->left->next = root->right;  
7:      if(root->right !=NULL)  
8:        root->right->next = root->next? root->next->left:NULL;  
9:       connect(root->left);  
10:       connect(root->right);  
11:    }