【Leetcode】Populating Next Right Pointers in Each Node

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

分析：对于树的处理，用的最多的就是遍历以及递归的算法，考虑的时候可以尽量的往这方面去靠；

递归的将左右子树跟他的右边的指针相连

/**
 * 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)
            return;
            
        connectToright(root, NULL);
        
    }
    
private:

    void connectToright(TreeLinkNode *node, TreeLinkNode *sibling) {
        if(node == NULL)
            return;
            
        node->next = sibling;
        
        if(node->left && node->right)
            connectToright(node->left, node->right);
        
        if(sibling != NULL)
        {
            connectToright(node->right, sibling->left);
        }
        else
            connectToright(node->right, NULL);
    }

};