leetcode question 116、117: Populating Next Right Pointers in Each Node（Ⅱ）

问题：

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.
• Recursive approach is fine, implicit stack space does not count as extra space for this problem.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

分析：

这道题叫我们对一课完全二叉树（在117中是一般二叉树）进行操作，如果当前节点右边有节点，就将next指针指向它，否则就置为空。我是用两个list来完成的，一个用来存储当前层的节点，另一个用来存储下一层的节点（按顺序），当前层处理完后，就将下一层的节点赋给当前层，一直到叶子节点。

 

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) {
        //为空直接跳出
        if(root == NULL){
            return;
        }
        
        //当前层
        list<TreeLinkNode*> current;
        TreeLinkNode* cur_node;//当前节点
        TreeLinkNode* temp_node;//右边的节点（没有时为空）
        
        current.push_back(root);
        
        while( true ){
            list<TreeLinkNode*> temp;
            
            while( current.size() != 0 ){
                cur_node = current.front();
                current.pop_front();

                if(cur_node->left != NULL){
                    temp.push_back(cur_node->left);
                }
                
                if(cur_node->right != NULL){
                    temp.push_back(cur_node->right);
                }

                //当前节点右边还有节点
                if(current.size()!=0){
                    temp_node = current.front();
                    cur_node->next = temp_node;
                }else{
                    cur_node->next = NULL;
                }
            }
           
            current = temp;
            
            if( current.size() == 0 && temp.size() == 0 ){
                break;
            }
        }
    }
};