Binary Tree Level Order Traversal (Easy)

Description

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

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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]
]

Analysis

dfs遍历一下，标记层数即可

My Solution

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//C++
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {

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