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

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 toNULL.

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,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

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:    }