MAY-24 2020 Challenge

Construct Binary Search Tree from Preorder Traversal

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements.

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Constraints:

1 <= preorder.length <= 100

1 <= preorder[i] <= 10^8

The values of preorder are distinct

Solution in Java:

class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if(preorder == null || preorder.length == 0){
            return null;
        }

        TreeNode root = new TreeNode(preorder[0]);
        Stack<TreeNode> stk = new Stack<TreeNode>();
        stk.push(root);
        for(int i = 1; i<preorder.length; i++){
            TreeNode cur = new TreeNode(preorder[i]);
            TreeNode top = stk.peek();

            while(!stk.isEmpty() && stk.peek().val<preorder[i]){
                top = stk.pop();
            }

            if(top.val < preorder[i]){
                top.right = cur;
            }else{
                top.left = cur;
            }
    stk.push(cur);
        }
  return root;
    }
}