[LeetCode]Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

这道题是对给定的一棵二叉树，要求分层遍历，即从上到下、从左至右的顺序访问每个结点并保存。这道题的一个注意点在于，不同层的结点保存在不同的vector容器中。所以，设立两个变量分别保存当前层结点个数curCount和下一层结点的个数nextCount。遍历当前层结点时，每访问一个结点，curCount减一，同时计算当前结点的左右孩子个数n，并nextCount+=n。当curCount减为0时表示当前层已经遍历结束，并将nextCount的值赋给curCount，nextCount清零。重复上述操作，直至队列为空。
下面贴上代码：

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int> > levelOrder(TreeNode *root) {        vector<vector<int> > ans;        queue<TreeNode*> q;        if (root == NULL)            return ans;        q.push(root);        int nextCount = 0;        int curCount = 1;        vector<int> temp;        while(!q.empty()){            TreeNode* p = q.front();            q.pop();            temp.push_back(p->val);            if (p->left){                q.push(p->left);                nextCount++;            }            if (p->right){                q.push(p->right);                nextCount++;            }            curCount--;            if (!curCount){                ans.push_back(temp);                temp=vector<int>();                curCount = nextCount;                nextCount = 0;            }        }        return ans;    }};