第三周：[LeetCode]102. Binary Tree Level Order Traversal

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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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]

其实就是对二叉树进行一遍广度优先搜索，每一层输出的结果保存为一个数组，用队列实现的时候维护count和next_count两个变量记录当前层需要遍历节点的个数以及下一层需要遍历节点的个数。

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
       vector<vector<int>> ans;
        vector<int> level;
        if(root == NULL)
            return ans;
        queue<TreeNode*> q;
        q.push(root);
        int count = 1,next_count = 0;
        while(!q.empty()){
            TreeNode* node = q.front();
            level.push_back(node -> val);
            if(node -> left != NULL){
                q.push(node -> left);
                next_count ++;
            }
            if(node -> right != NULL){
                q.push(node -> right);
                next_count ++;
            }
            q.pop();
            count --;
            if(count == 0){
                ans.push_back(level);
                level.clear();
                count = next_count;
                next_count = 0;
            }
        }
        return ans;
    }
};