102. Binary Tree Level Order Traversal

问题描述：

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]]

解题思路：

层次二叉树的输出，用dfs，利用一个vector[level]记录level层的数值并输出：

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ret;
        if(root==NULL) return ret;
        dfs(root,0,ret);
        return ret;
    }
    void dfs(TreeNode *root,int level,vector<vector<int>> &ret)
    {
        if(root==NULL) return;
        if(ret.size()==level) ret.push_back({});
        ret[level].push_back(root->val);
        dfs(root->left,level+1,ret);
        dfs(root->right,level+1,ret);
    }
};