Last updated on October 5th, 2024 at 04:40 pm

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

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