LeetCode OJ-- Populating Next Right Pointers in Each Node II **@

https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

接上一题目，输入的树不是perfect的。

在这里深搜就不好使了。因为，在你往下深搜的时候，可能上一行的next还没处理到那里，所以会丢失信息。

利用它的next结构，进行bfs，同时记录下一行的head节点。

/**
 * 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;
        
        TreeLinkNode *NextRowHead = NULL;
        TreeLinkNode *NextRowPrev = NULL;
        while(root)
        {
            if(root->left)
            {
                if(NextRowHead == NULL) // the first element of next row
                {
                    NextRowHead = root->left;
                }
                else
                    NextRowPrev->next = root->left; //它前面有点
                    
                NextRowPrev = root->left;
            }
            if(root->right)
            {
                if(NextRowHead == NULL)
                {
                    NextRowHead = root->right;
                }
                else
                    NextRowPrev->next = root->right;
                    
                NextRowPrev = root->right;
            }
            root = root->next;
            
            if(root == NULL) //begin a new row
            {
                root = NextRowHead;
                NextRowHead = NULL;
                NextRowPrev = NULL;
            }
        }
    } 
};