一天一道LeetCode

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（一）题目

来源：

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

（二）解题

题目大意：定义一个新的二叉树结构，增加一个新指针next指向该节点右侧相邻的节点。

解题思路：考虑到要指向右侧相邻的节点，可以采用层序遍历的方式来求解

将每一层的节点用next连接起来即可！

层序遍历可以参考：

废话不说，看AC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;        queue
   
     myque;        myque.push(root);        while(!myque.empty())        {            queue
    
      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        }    }};

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