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In this post, you will find the solution for the **Binary Tree Level Order Traversal II** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Binary Tree Level Order Traversal II– LeetCode Problem

Binary Tree Level Order Traversal II– LeetCode Problem

**Problem:**

Given the `root`

of a binary tree, return *the bottom-up level order traversal of its nodes’ values*. (i.e., from left to right, level by level from leaf to root).

**Example 1:**

Input:root = [3,9,20,null,null,15,7]Output:[[15,7],[9,20],[3]]

**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 II– LeetCode Solutions

Binary Tree Level Order Traversal II Solution in C++:

class Solution { public: vector> levelOrderBottom(TreeNode* root) { if (!root) return {}; vector > ans; queue q{{root}}; while (!q.empty()) { vector currLevel; for (int size = q.size(); size > 0; --size) { TreeNode* node = q.front(); q.pop(); currLevel.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } ans.push_back(currLevel); } reverse(begin(ans), end(ans)); return ans; } };

Binary Tree Level Order Traversal II Solution in Java:

class Solution { public List> levelOrderBottom(TreeNode root) { if (root == null) return new ArrayList<>(); List

> ans = new ArrayList<>(); Queue

q = new LinkedList<>(Arrays.asList(root)); while (!q.isEmpty()) { List currLevel = new ArrayList<>(); for (int size = q.size(); size > 0; --size) { TreeNode node = q.poll(); currLevel.add(node.val); if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } ans.add(currLevel); } Collections.reverse(ans); return ans; } }

Binary Tree Level Order Traversal II Solution in Python:

class Solution: def levelOrderBottom(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: return [] ans = [] q = deque([root]) while q: currLevel = [] for _ in range(len(q)): node = q.popleft() currLevel.append(node.val) if node.left: q.append(node.left) if node.right: q.append(node.right) ans.append(currLevel) return ans[::-1]