Binary Tree Level Order Traversal II LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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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<vector<int>> levelOrderBottom(TreeNode* root) {
    if (!root)
      return {};

    vector<vector<int>> ans;
    queue<TreeNode*> q{{root}};

    while (!q.empty()) {
      vector<int> 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<List<Integer>> levelOrderBottom(TreeNode root) {
    if (root == null)
      return new ArrayList<>();

    List<List<Integer>> ans = new ArrayList<>();
    Queue<TreeNode> q = new LinkedList<>(Arrays.asList(root));

    while (!q.isEmpty()) {
      List<Integer> 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]