Leetcode Binary Tree Level Order Traversal problem solution YASH PAL, 31 July 202419 January 2026 In this Leetcode Binary Tree Level Order Traversal problem solution, we have given the root of a binary tree, return the level order traversal of its nodes’ values.Leetcode Binary Tree Level Order Traversal problem solution in Python.class Solution: def levelOrder(self, root: TreeNode) -> List[List[int]]: if not root: return None queue = [root] result = [] while queue: arr = [] for i in range(len(queue)): node = queue.pop(0) arr.append(node.val) left = node.left right = node.right if left: queue.append(left) if right: queue.append(right) result.append(arr) Binary Tree Level Order Traversal problem solution in Java.public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if(root == null){ return res; } Queue<TreeNode> q = new LinkedList(); q.add(root); while(!q.isEmpty()){ int size = q.size(); List<Integer> level = new ArrayList<>(); while(size > 0){ TreeNode node = q.poll(); level.add(node.val); if(node.left != null){ q.add(node.left); } if(node.right != null){ q.add(node.right); } size--; } res.add(level); } return res; } Problem solution in C++.class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { queue<TreeNode *> q; vector<vector<int>> v; vector<int> tmp; if (!root) return {}; tmp.push_back(root->val); q.push(root); v.push_back(tmp); while (!q.empty()) { int size = q.size(); vector<int> tmp; for (int i = 0; i < size; i++) { TreeNode *n = q.front(); q.pop(); //cout << n->val << endl; if (n->left) { tmp.push_back(n->left->val); q.push(n->left); } if (n->right) { tmp.push_back(n->right->val); q.push(n->right); } } if (!tmp.empty()) v.push_back(tmp); } return v; } }; Problem solution in C.void addnum(int size, int **array, int val, int level) { array[level] = realloc(array[level], size * sizeof (int)); array[level][size-1] = val; } int maxdepth(struct TreeNode* root) { if (root == NULL) { return (0); } int ldepth = maxdepth(root->left) + 1; int rdepth = maxdepth(root->right) + 1; if (ldepth > rdepth) return ldepth; else return rdepth; } void currentlevel(struct TreeNode *root, int *returnColumnSizes, int **array, int level) { if (root == NULL) { return; } returnColumnSizes[level] += 1; addnum(returnColumnSizes[level], array, root->val, level); level++; currentlevel(root->left, returnColumnSizes, array, level); currentlevel(root->right, returnColumnSizes, array, level); } int** levelOrder(struct TreeNode* root, int* returnSize, int** returnColumnSizes){ int depth, **array; depth = maxdepth(root); *returnSize = depth; *returnColumnSizes = calloc(depth, sizeof(int)); array = calloc(depth, sizeof (int *)); currentlevel(root, *returnColumnSizes, array, 0); return (array); } coding problems solutions Leetcode Problems Solutions Leetcode