Convert Sorted List to Binary Search Tree in C++

Suppose we have a singly linked list where elements are sorted in ascending order, we have to convert it to a height balanced BST. So if the list is like [-10, -3, 0, 5, 9], The possible tree will be like −

To solve this, we will follow these steps −

• If the list is empty, then return null

• Define a recursive method called sortedListToBST() this will take list start node

• x := address of the previous node of mid node from list a

• mid := exact mid node

• create a new node with value by taking from the value of mid

• nextStart := next of mid node

• set mid next as null

• right of node := sortedListToBST(nextStart)

• if x is not null, then next of x = null and left of node := sortedListToBST(a)

• return node

Example (C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class ListNode{
   public:
   int val;
   ListNode *next;
   ListNode(int data){
      val = data;
      next = NULL;
   }
};
ListNode *make_list(vector<int> v){
   ListNode *head = new ListNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      ListNode *ptr = head;
      while(ptr->next != NULL){
         ptr = ptr->next;
      }
      ptr->next = new ListNode(v[i]);
   }
   return head;
}
class TreeNode{
   public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = right = NULL;
   }
};
void inord(TreeNode *root){
   if(root != NULL){
      inord(root->left);
      cout << root->val << " ";
      inord(root->right);
   }
}
class Solution {
   public:
   pair <ListNode*, ListNode*> getMid(ListNode* a){
      ListNode* prev = NULL;
      ListNode* fast = a;
      ListNode* slow = a;
      while(fast && fast->next){
         fast = fast->next->next;
         prev = slow;
         slow = slow->next;
      }
      return {prev, slow};
   }
   TreeNode* sortedListToBST(ListNode* a) {
      if(!a)return NULL;
      pair<ListNode*, ListNode*> x = getMid(a);
      ListNode* mid = x.second;
      TreeNode* Node = new TreeNode(mid->val);
      ListNode* nextStart = mid->next;
      mid->next = NULL;
      Node->right = sortedListToBST(nextStart);
      if(x.first){
         x.first->next = NULL;
         Node->left = sortedListToBST(a);
      }
      return Node;
   }
};
main(){
   vector<int> v = {-10,-3,0,5,9};
   ListNode *head = make_list(v);
   Solution ob;
   inord(ob.sortedListToBST(head));
}

Input

[-10,-3,0,5,9]

Output

-10 -3 0 5 9

Arnab Chakraborty

Published on 31-Mar-2020 11:58:45