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;
}

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