leetcode 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

 

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• Total Accepted: 118503
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• Difficulty: Easy

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