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

思路：

递归和非递归的思路差不多，简单说下递归吧：

根节点的位置比较特殊，不需要往右边链接的，但是到了下面，就要判断一下额外的条件了，一个是自身右边的链接是否打通还有一个就是自身是否有右节点，没有直接返回，有的话就是自身右节点孩子的next 指向自身 next 的左孩子。

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)  return ;
        if(root->left&&root->right){
            root->left->next=root->right;
        }
        if(root->next&&root->left){//到了非根节点的时候
            root->right->next=root->next->left;
        }
        
        connect(root->left);connect(root->right);
    }
*/    
    void connect(TreeLinkNode *root){
        if(root==NULL)  return ;
        
        while(root&&root->left){
            TreeLinkNode *tmp=root->left;
            while(root){
                root->left->next=root->right;
                if(root->next){
                    root->right->next=root->next->left;
                }
                root=root->next;
            }
            root=tmp;
        }
    }
};