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;
}