In this post, we are going to solve the 107. Binary Tree Level Order Traversal II problem of Leetcode. This problem 107. Binary Tree Level Order Traversal II is a Leetcode medium level problem. Let’s see the code, 107. Binary Tree Level Order Traversal II – Leetcode Solution.

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

Now, let’s see the code of 107. Binary Tree Level Order Traversal II – Leetcode Solution.

Binary Tree Level Order Traversal II – Leetcode Solution

107. Binary Tree Level Order Traversal II – Solution in Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        Stack<List<Integer>> s = new Stack<>();
       if(root == null) return ans;
        Queue<TreeNode> q = new LinkedList<>();
        q.add(root);
        while(!q.isEmpty()){
            int size = q.size();
            List<Integer> level = new ArrayList<Integer>();
            while(size-- > 0){
                TreeNode rn = q.poll();
                level.add(rn.val);
                if(rn.left != null) q.add(rn.left);
                if(rn.right != null) q.add(rn.right);
            }
             
            ans.add(0,level);
        }
        return ans;
    }
}

107. Binary Tree Level Order Traversal II – Solution in C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int height(TreeNode* root){
        if(root == NULL) return 0;
        
        return max(height(root->left),height(root->right)) + 1;
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        int n = height(root);
        vector<vector<int>> ans(n);
        if(root == NULL) return ans;
        
        queue<TreeNode*>qu;
        qu.push(root);
        while(!qu.empty()){
            int sz = qu.size();
            vector<int>level;
            for(int i=0;i<sz;i++){
                TreeNode* front = qu.front();
                qu.pop();
                level.push_back(front->val);
                if(front->left) qu.push(front->left);
                if(front->right) qu.push(front->right);
            }
            ans[n-1] = level;
            n--;
        }
        return ans;
    }
};

107. Binary Tree Level Order Traversal II – Solution in Python

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def levelOrderBottom(self, root):
        queue, res = collections.deque([(root, 0)]), []
        while queue:
            node, level = queue.popleft()
            if node:
                if len(res) < level+1:
                    res.insert(0, [])
                res[-(level+1)].append(node.val)
                queue.append((node.left, level+1))
                queue.append((node.right, level+1))
        return res
        

Note: This problem 107. Binary Tree Level Order Traversal II is generated by Leetcode but the solution is provided by CodingBroz. This tutorial is only for Educational and Learning purpose.