leetcode || 116、Populating Next Right Pointers in Each Node

problem：

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

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  Tree Depth-first Search

题意：为二叉树的每一个节点增加一个next指针，将二叉树的每层节点链接起来

thinking：

（1）类似于B*树，每层节点也链接起来。

（2）看到题，我立即想到用层序遍历来做，采用BFS，但是题目要求constant extra space，目的是想用DFS来做做，
由于这道题的二叉树的特殊性（满树）不具有普遍意义。所以我采用更具普遍意义的BFS方法，可能空间复杂度不满足要求，但也AC了

（3）BFS  是采用queue结构存储每一层的节点。

code：

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root==NULL)
            return;
        queue<TreeLinkNode*> queue0;
        queue0.push(root);
        level_visit(queue0);
        return;
    }
protected:
    void level_visit(queue<TreeLinkNode*> queue1)
    {
        if(queue1.empty())
            return;
        queue<TreeLinkNode*> queue2=queue1;
        queue<TreeLinkNode*> queue3;
        TreeLinkNode *tmp=queue1.front();
        queue1.pop();
        tmp->next=NULL;
        while(!queue1.empty())
        {
            TreeLinkNode *tmp2=queue1.front();
            queue1.pop();
            tmp2->next=NULL;
            tmp->next=tmp2;
            tmp=tmp2;
        }
        while(!queue2.empty())
        {
            TreeLinkNode *node=queue2.front();
            queue2.pop();
            if(node->left!=NULL && node->right!=NULL)
            {
                queue3.push(node->left);
                queue3.push(node->right);
            }
        }
        level_visit(queue3);
    }
    
};