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Leetcode Binary Tree Level Order Traversal II problem solution

YASH PAL, 31 July 2024

In this Leetcode Binary Tree Level Order Traversal II problem solution we have Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values.

Leetcode Binary Tree Level Order Traversal II problem solution

Problem solution in Python.

class Solution:
    def levelOrderBottom(self, root):
        result = []
        if root == None:
            return result
        def bfsbottomup(result, prev):
            if prev == []:
                return
            temp = []
            cur = []
            for node in prev:
                temp.append(node.val)
                if node.left != None:
                    cur.append(node.left)
                if node.right != None:
                    cur.append(node.right)
            bfsbottomup(result, cur)
            result.append(temp)
        bfsbottomup(result, [root])
        return result

Problem solution in Java.

class Solution {
    LinkedList ans = new LinkedList();
    Queue<TreeNode> q = new ArrayDeque();
    
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        
        if(root == null) return ans;
        
        q.offer(root);
        while(!q.isEmpty()){
            
            List <Integer>tmp = new ArrayList();
            
            int size = q.size();
            for(int i = 0 ; i < size; i++){
                
                TreeNode node = q.poll();
                tmp.add(node.val);
                
                if(node.left != null){
                    q.add(node.left);   
                }
                if(node.right != null){
                    q.add(node.right);
                }
            }
            ans.addFirst(tmp);
        }
        
        return ans;
    }
}

Problem solution in C++.

vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> result;
        queue<pair<TreeNode*,int>> toVisit;
        toVisit.push(make_pair(root,0));
        while(!toVisit.empty()) {
            TreeNode *atNode = toVisit.front().first;
            int level = toVisit.front().second;
            toVisit.pop();
            
            if(!atNode) {
                continue;
            }

            if(result.size() <= level) {
                vector<int> empty;
                result.push_back(empty);
            }
            
            result[level].push_back(atNode->val);
            toVisit.push(make_pair(atNode->left,level+1));
            toVisit.push(make_pair(atNode->right,level+1));
        }
        for(int i=0; i<result.size()/2; i++) {
            vector<int> temp = result[i];
            result[i] = result[result.size()-i-1];
            result[result.size()-i-1] = temp;
        }
        return result;
    }

Problem solution in C.

int maxDepth(struct TreeNode* root) {
    if (root==NULL) return 0;
    else {
        int L=maxDepth(root->left);
        int R=maxDepth(root->right);
        int M=(L>R)?L:R;
        
        return M+1;
    }
}

int** levelOrderBottom(struct TreeNode* root, int** columnSizes, int* returnSize) {
    if (root==NULL) return NULL;
    int depth=*returnSize=maxDepth(root);
    
    int** res=(int**)calloc(depth,sizeof(int*));
    *columnSizes=(int*)calloc(depth,sizeof(int));
    
    struct TreeNode* queue[5000];
    int front=0,rear=0;
    queue[rear++]=root;
    int cur_size=1;int next_size=0;int size=depth-1;

    while (front<rear) {
        res[size]=(int*)malloc(2000*sizeof(int));
        
        for (int i=0;i<cur_size&&front<rear;i++) {
            struct TreeNode* temp=queue[front++];
            res[size][i]=temp->val;
            
            if (temp->left) {
                queue[rear++]=temp->left;
                next_size++;
            }
            if (temp->right) {
                queue[rear++]=temp->right;
                next_size++;
            }
        }
        
        (*columnSizes)[size--]=cur_size;

        cur_size=next_size;
        if (cur_size==0) break;
        next_size=0;
    }
    
    return res;
}

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