leetcode 102. Binary Tree Level Order Traversal

102. Binary Tree Level Order Traversal

 

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• Total Accepted: 118503
• Total Submissions: 343625
• 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;
    }
};