Populating Next Right Pointers in Each Node

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

分析：这道题就是层次遍历，把每一层的结点链接成单向链表，而且允许我们使用额外空间来解题，故而我们可以使用队列很容易就可以把此题解出来。每次把每一层的结点存入队列，然后把头结点取出，其结点的next指向队列新的头结点，当这一层结束时，插入NULL。循环进行，就可以了。
C++代码片段：

/**
 * 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) {
        queue<TreeLinkNode*> qroot;
        if(root==NULL)
            return;
        qroot.push(root);
        qroot.push(NULL);
        while(qroot.size())
        {
            TreeLinkNode* currentNode=qroot.front();
            qroot.pop();
            if(currentNode)
            {
                currentNode->next=qroot.front();
                if(currentNode->left)
                    qroot.push(currentNode->left);
                if(currentNode->right)
                    qroot.push(currentNode->right);
            }
            else
            {
                if(qroot.size())
                {
                    qroot.push(NULL);
                }
            }
            
        }
    }
};