Binary Tree Level Order Traversal LeetCode Solution

Problem – Binary Tree Level Order Traversal LeetCode Solution

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

Binary Tree Level Order Traversal LeetCode Solution in Java

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        
        if(root == null) return wrapList;
        
        queue.offer(root);
        while(!queue.isEmpty()){
            int levelNum = queue.size();
            List<Integer> subList = new LinkedList<Integer>();
            for(int i=0; i<levelNum; i++) {
                if(queue.peek().left != null) queue.offer(queue.peek().left);
                if(queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(subList);
        }
        return wrapList;
    }
}

Binary Tree Level Order Traversal LeetCode Solution in C++

vector<vector<int>> ret;

void buildVector(TreeNode *root, int depth)
{
    if(root == NULL) return;
    if(ret.size() == depth)
        ret.push_back(vector<int>());
    
    ret[depth].push_back(root->val);
    buildVector(root->left, depth + 1);
    buildVector(root->right, depth + 1);
}

vector<vector<int> > levelOrder(TreeNode *root) {
    buildVector(root, 0);
    return ret;
}

Binary Tree Level Order Traversal LeetCode Solution in Python

def levelOrder(self, root):
    ans, level = [], [root]
    while root and level:
        ans.append([node.val for node in level])
        LRpair = [(node.left, node.right) for node in level]
        level = [leaf for LR in LRpair for leaf in LR if leaf]
    return ans

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