Last updated on July 18th, 2024 at 03:00 am

Here, We see Binary Tree Level Order Traversal LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

List of all LeetCode Solution

Topics

Breadth First Search, Tree

Companies

Amazon, Apple, Bloomberg, Facebook, LinkedIn, Microsoft

Level of Question

Medium

Binary Tree Level Order Traversal LeetCode Solution

Problem Statement

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: []

1. Binary Tree Level Order Traversal LeetCode Solution C++

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;
    }
};

2. Binary Tree Level Order Traversal LeetCode Solution Java

class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) 
    {
        List<List<Integer>>al=new ArrayList<>();
        pre(root,0,al);
        return al;
    }
    public static void pre(TreeNode root,int l,List<List<Integer>>al)
    {
        if(root==null)
            return;
        if(al.size()==l)
        {
            List<Integer>li=new ArrayList<>();
            li.add(root.val);
            al.add(li);
        }
        else
            al.get(l).add(root.val);
        pre(root.left,l+1,al);
        pre(root.right,l+1,al);
    } 
}

3. Binary Tree Level Order Traversal Solution JavaScript

var levelOrder = function(root) {
    let q = [root], ans = []
    while (q[0]) {
        let qlen = q.length, row = []
        for (let i = 0; i < qlen; i++) {
            let curr = q.shift()
            row.push(curr.val)
            if (curr.left) q.push(curr.left)
            if (curr.right) q.push(curr.right)
        }
        ans.push(row)            
    }
    return ans
};

4. Binary Tree Level Order Traversal Solution Python

class Solution(object):
    def levelOrder(self, root):
        if not root:
            return []
        Q = deque([root])
        levels = [[root.val]]
        temp = deque()
        while Q:
            node = Q.popleft()
            if node.left: temp.append(node.left)
            if node.right: temp.append(node.right)
            if not Q:
                if temp:
                    levels.append([n.val for n in temp])
                Q = temp
                temp = deque()
        return levels