Binary Tree Level Order Traversal II LeetCode Solution

Problem – Binary Tree Level Order Traversal II LeetCode Solution

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 Solution in Java

public class Solution {
    public List<List<Integer>> levelOrderBottom(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(0, subList);
        }
        return wrapList;
    }
}

Binary Tree Level Order Traversal II LeetCode Solution in Python

def levelOrderBottom1(self, root):
    res = []
    self.dfs(root, 0, res)
    return res

def dfs(self, root, level, res):
    if root:
        if len(res) < level + 1:
            res.insert(0, [])
        res[-(level+1)].append(root.val)
        self.dfs(root.left, level+1, res)
        self.dfs(root.right, level+1, res)

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

void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level) {
    if (!node) return;
    if (level >= ans.size())
        ans.push_back({});
    ans[level].push_back(node->val);
    levelOrder(ans,node->left,level+1);
    levelOrder(ans,node->right,level+1);
}

vector<vector<int>> levelOrderBottom(TreeNode* root) {
    vector<vector<int>> ans;
    levelOrder(ans,root,0);
    reverse(ans.begin(),ans.end());
    return ans;
}
Binary Tree Level Order Traversal II LeetCode Solution Review:

In our experience, we suggest you solve this Binary Tree Level Order Traversal II LeetCode Solution and gain some new skills from Professionals completely free and we assure you will be worth it.

If you are stuck anywhere between any coding problem, just visit Queslers to get the Binary Tree Level Order Traversal II LeetCode Solution

Find on Leetcode

Conclusion:

I hope this Binary Tree Level Order Traversal II LeetCode Solution would be useful for you to learn something new from this problem. If it helped you then don’t forget to bookmark our site for more Coding Solutions.

This Problem is intended for audiences of all experiences who are interested in learning about Data Science in a business context; there are no prerequisites.

Keep Learning!

More Coding Solutions >>

LeetCode Solutions

Hacker Rank Solutions

CodeChef Solutions

Leave a Reply

Your email address will not be published.