Problem description:

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],

1
2
3
4
5
  3
 / \
9  20
  /  \
 15   7

return its level order traversal as:

1
2
3
4
5
[
  [3],
  [9,20],
  [15,7]
]

Solution:

This is a very basic problem, we can use dfs to solve this problem. One thing to notice is that we need to use a depth variable to store the level where we are.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * 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> > levelOrder(TreeNode *root) {
        dfs(root, 0);
        return res;
    }
    
    void dfs(TreeNode *root, int depth)
    {
        if(root == NULL) return;
        if(res.size() == depth) //need to insert a new layer
            res.push_back(vector<int>());

        res[depth].push_back(root->val);
        dfs(root->left, depth + 1);
        dfs(root->right, depth + 1);
    }
private:
    vector<vector<int>> res;

};

time complexity: O(n)

Second time:

Use BFS to do traversal

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
/**
 * 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>> levelOrder(TreeNode* root) {
        if(!root) return vector<vector<int>>{};
        vector<vector<int>> res;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            res.push_back(vector<int>{});
            int size= q.size();
            for(int i= 0; i< size; i++){
                TreeNode* tmp= q.front(); q.pop();
                res[res.size()-1].push_back(tmp->val);
                if(tmp->left) q.push(tmp->left);
                if(tmp->right) q.push(tmp->right);
            }
        }
        return res;
    }
};

time complexity: $O(n)$
space complexity: $O(n)$