leetcode刷题，总结，记录，备忘。116

leetcode116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.
• 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

使用c++队列容器的话这个题目很简单。一层一层的进行处理，用count计数统计每一层的节点个数。将每个节点的next指向同一层右边的节点，最后一个指向null。

/**
 * 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*> qTree;
        int count = 1;
        if (!root)
        return;
        qTree.push(root);
        while (!qTree.empty())
        {
            TreeLinkNode * temp = qTree.front();
            qTree.pop();
            if (temp->left)
            qTree.push(temp->left);
            if (temp->right)
            qTree.push(temp->right);
            count--;
            if (count == 0)
            {
            temp->next = NULL;
            count = qTree.size();
            }
            else
            {
                temp->next = qTree.front();
            }
        }
        return;
    }
};