【一天一道LeetCode】#117. Populating Next Right Pointers in Each Node II

一天一道LeetCode

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（一）题目

Follow up for problem “Populating Next Right Pointers in Each Node”.

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,

Given the following binary tree,

    1
  /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

    1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \    \
4-> 5 -> 7 -> NULL

（二）解题

参考：【一天一道LeetCode】#116. Populating Next Right Pointers in Each Node

没想到昨天做的解法直接把今天的题给解了。一模一样的代码，请跳转到上题的解题思路：

/**
 * 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;
        queue<TreeLinkNode *> myque;
        myque.push(root);
        while(!myque.empty())
        {
            queue<TreeLinkNode *> temp_que;
            TreeLinkNode *pre = NULL;
            while(!myque.empty())
            {
                TreeLinkNode *temp = myque.front();
                myque.pop();
                if(pre==NULL) pre = temp;//相当于每一层的头节点
                else {
                    pre->next = temp;//用next指针连接每一层的节点
                    pre = temp;
                }
                if(temp->left!=NULL) temp_que.push(temp->left);//下一层
                if(temp->right!=NULL) temp_que.push(temp->right);//下一层
            }
            pre->next=NULL;//最后一个节点的next需要指向NULL
            myque=temp_que;//进入下一层操作
        }
    }
};