题解 | #求二叉树的层序遍历#

用队列实现，每一层的节点的个数就是que的元素的个数，没保存一个节点的值，同时将相应节点的左右子结点保存进去。

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 * };
 */

class Solution {
public:
    /**
     * 
     * @param root TreeNode类 
     * @return int整型vector<vector<>>
     */
    vector<vector<int> > levelOrder(TreeNode* root) {
        // write code here
        vector<vector<int>> result{};
        queue<TreeNode*> que{};
        if(root==nullptr) return result;
        que.push(root);
        while(!que.empty()){
            int n=que.size();
            vector<int> level{};
            for(int i=0;i<n;++i){
                auto cur=que.front();
                que.pop();
                level.push_back(cur->val);
                if(cur->left) que.push(cur->left);
                if(cur->right) que.push(cur->right);
            }
            result.push_back(level);
        }
        return result;
    }
};