LeetCode - Construct Binary Search Tree from Preorder Traversal - 30Days Challenge

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] 8 / \ 5 10 / \ \ 1 7 12

Constraints:

1 <= preorder.length <= 100
1 <= preorder[i] <= 10^8
The values of preorder are distinct.

Code


 /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
   int i = 0;
    public TreeNode bstFromPreorder(int[] A) {
        return bstFromPreorder(A, Integer.MAX_VALUE);
    }

    public TreeNode bstFromPreorder(int[] A, int bound) {
        if (i == A.length || A[i] > bound) return null;
        TreeNode root = new TreeNode(A[i++]);
        root.left = bstFromPreorder(A, root.val);
        root.right = bstFromPreorder(A, bound);
        return root;
    }
}