Level Order Binary Tree to Linked List - Amazon Top Interview Questions


Problem Statement :


Given a binary tree root, convert it to a singly linked list using level order traversal.

Constraints

1 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [2, null, null], [3, [4, null, null], [5, null, null]]]

Output

[1, 2, 3, 4, 5]

Example 2

Input

root = [1, [0, null, null], [2, null, null]]

Output

[1, 0, 2]



Solution :



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                        Solution in C++ :

void rec(Tree *root, int h, vector<LLNode *> &t) {
    if (!root) return;
    LLNode *tmp = new LLNode();
    tmp->val = root->val;
    if (t.size() >= h + 1) {
        tmp->next = t[h];
        t[h] = tmp;
    } else {
        tmp->next = nullptr;
        t.push_back(tmp);
    }
    rec(root->right, h + 1, t);
    rec(root->left, h + 1, t);
}
LLNode *solve(Tree *root) {
    vector<LLNode *> t;
    rec(root, 0, t);
    LLNode *h, *dummy = new LLNode();
    h = dummy;
    for (LLNode *cur : t)
        while (cur) {
            dummy->next = cur;
            cur = cur->next;
            dummy = dummy->next;
        }
    return h->next;
}
                    




                        Solution in Python : 
                            
class Solution:
    def solve(self, root):
        Head = LL = LLNode(None)
        d = deque()
        d.append(root)
        while d:
            current = d.popleft()
            LL.next = LLNode(current.val)
            LL = LL.next
            if current.left:
                d.append(current.left)
            if current.right:
                d.append(current.right)
        return Head.next
                    


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