Skip to content

102. Binary Tree Level Order Traversal

Difficulty Topics

Description

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

binary-tree-level-order-traversal.py
# 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:
    def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root: return []

        res = []
        dq = deque([root])

        while dq:
            n = len(dq)
            curr = []

            for _ in range(n):
                node = dq.popleft()
                curr.append(node.val)

                for child in (node.left, node.right):
                    if child:
                        dq.append(child)

            res.append(curr)

        return res
binary-tree-level-order-traversal.cpp
/**
 * 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) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> q; q.push(root);
        vector<vector<int>> res;

        if (root == nullptr) return res;

        while (q.size() > 0){
            vector<int> tmp;
            int length = q.size();

            while (length--){
                TreeNode* node = q.front();
                tmp.push_back(node->val);
                q.pop();

                if (node->left){
                    q.push(node->left);
                }

                if (node->right){
                    q.push(node->right);
                }
            }

            res.push_back(tmp);
        }

        return res;
    }
};