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

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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:

tree1
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]

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

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  29. Norma ISO 10816
    Sistemas de ajuste: clave para el desempeño fluido y eficiente de las dispositivos.

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    Reply

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