Binary Tree Level Order Traversal

Problem

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]

Example 2:

Input: root = [1]
Output: [[1]]

Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

Solution

같은 레벨의 노드를 출력한다
left/right의 값이 있으면, q에 차례로 쌓는다
쌓은 q를 반환하기 위한 ans vector에 넣어준다

Code

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
####  *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
#include <vector>

class Solution
{
public:
    vector<vector<int>> levelOrder(TreeNode *root)
    {
        vector<vector<int>> ans;
        if (root == NULL)
            return ans;
        queue<TreeNode *> q;
        q.push(root);
        while (!q.empty())
        {
            int s = q.size();
            vector<int> v;
            for (int i = 0; i < s; i++)
            {
                TreeNode *node = q.front();
                q.pop();
                if (node->left != NULL)
                    q.push(node->left);
                if (node->right != NULL)
                    q.push(node->right);
                v.push_back(node->val);
            }
            ans.push_back(v);
        }
        return ans;
    }
};

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

class Solution:
    # BFS Iterative solution 
    # Time complexity - O(n) | Space complexity - O(n)
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return root
        res = []
        nodes = deque([root])
        while nodes:
            l = len(nodes)
            temp = []
            for i in range(l):
                node = nodes.popleft()
                if node:
                    temp.append(node.val)
                    nodes.append(node.left)
                    nodes.append(node.right)
           
            if len(temp)>0:
                res.append(temp)
        return res