【LeetCode笔记】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

思路：题目中说明 You may only use constant extra space. 所以不能使用递归啦~改用指针&双重循环

1.从最右节点开始走（图中1，2，4）

2.如果当前节点是上个节点的左节点，那么其next是上个节点的右节点

3.如果当前节点是三个节点的右节点，那么其next是上个结点的next的左节点。

4.循环中注意判断是否为空。

/**
 * 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;
        while(root->left!=NULL){
            TreeLinkNode* cur = root;
            while(cur!=NULL){
                cur->left->next = cur->right;
                if(cur->next!=NULL){
                    cur->right->next = cur->next->left;
                }
                cur = cur->next;
            }
            root = root->left;
        }
    }
};