【leetcode】102. Binary Tree Level Order Traversal

按照广度优先的思想，设置两个队列，一个队列存放当前level的所有节点，另外一个队列存放当前节点的子节点。

/**
 * @author          johnsondu
 * @problem         Binary Tree Level Order Traversal
 * @url             https://leetcode.com/problems/binary-tree-level-order-traversal/
 * @timeComlexity   O(n)
 * @spaceComplexity O(n)
 * @strategy        bfs
 */

/**
 * Definition for a binary tree node.
 * 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*> q1, q2;
        if(root == NULL) return ans;
        
        q1.push(root);
        vector<int> layer;
        
        while(!q1.empty()) {
            layer.clear();
            while(!q1.empty()) {
                TreeNode* cur = q1.front();
                q1.pop();
                layer.push_back(cur->val);
                if(cur->left) {
                    q2.push(cur->left);
                }
                if(cur->right) {
                    q2.push(cur->right);
                }
            }
            ans.push_back(layer);
            q1 = q2;
            while(!q2.empty()) q2.pop();
        }
        
        return  ans;
    }
};