【LeetCode】116. Populating Next Right Pointers in Each Node

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,

         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

提示：

题目的关键就是找规律：

• 左子节点的next必定是右子节点；
• 如果父节点的next是NULL，则右子节点的next也是NULL；
• 如果父节点的next不是NULL，则右子节点的next是父节点的next节点的左子节点。

代码：

/**
 * 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 || root->left == NULL)
            return;
        root->left->next = root->right;
        if (root->next != NULL) 
            root->right->next = root->next->left;
        connect(root->left);
        connect(root->right);
    }
};