Make Binary Tree From Linked List

Problem Statement - link #

Given a Linked List Representation of Complete Binary Tree. The task is to construct the Binary tree. Note : The complete binary tree is represented as a linked list in a way where if root node is stored at position i, its left, and right children are stored at position 2*i+1, 2*i+2 respectively.

Your Task: The task is to complete the function convert() which takes head of linked list and root of the tree as the reference. The driver code prints the level order.

Expected Time Complexity: O(N) Expected Auxiliary Space: O(N)

Examples #

Example 1:

Input:
N = 5
K = 1->2->3->4->5
Output: 1 2 3 4 5
Explanation: The tree would look like
      1
    /   \
   2     3
 /  \
4   5
Now, the level order traversal of
the above tree is 1 2 3 4 5.

Example 2:

Input:
N = 5
K = 5->4->3->2->1
Output: 5 4 3 2 1
Explanation: The tree would look like
     5
   /  \
  4    3
 / \
2    1
Now, the level order traversal of
the above tree is 5 4 3 2 1..

Constraints #

• 1 <= Number of nodes <= 10^5
• 0 <= Data of a node <= 10^5

Solutions #

//Function to make binary tree from linked list.
TreeNode *bt(vector<int> &v, int i){
    if(i>=v.size()) return NULL;
    TreeNode* root = new TreeNode(v[i]);
    root->left = bt(v,2*i+1);
    root->right = bt(v,2*i+2);
    return root;
}
void convert(Node *head, TreeNode *&root) {
    // Your code here
    vector<int> v;
    Node* temp = head;
    while(temp) { v.push_back(temp->data); temp=temp->next; }
    root = bt(v,0);
}