1008.construct-binary-search-tree-from-preorder-traversal.java

/*
 * @lc app=leetcode id=1008 lang=java
 *
 * [1008] Construct Binary Search Tree from Preorder Traversal
 *
 * https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/description/
 *
 * algorithms
 * Medium (75.48%)
 * Likes:    879
 * Dislikes: 30
 * Total Accepted:    81.5K
 * Total Submissions: 106.3K
 * Testcase Example:  '[8,5,1,7,10,12]'
 *
 * 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.)
 * 
 * 
 * 
 * Example 1:
 * 
 * 
 * Input: [8,5,1,7,10,12]
 * Output: [8,5,10,1,7,null,12]
 * 
 * 
 * 
 * 
 * 
 * Note: 
 * 
 * 
 * 1 <= preorder.length <= 100
 * The values of preorder are distinct.
 * 
 * 
 */

// @lc code=start
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        return build(preorder, 0, preorder.length - 1);
    }
    
    private TreeNode build(int[] preorder, int s, int e) {
        
        if (s > e) return null;
        TreeNode root = new TreeNode(preorder[s]);
       
        int index = s + 1;
        while (index <= e && preorder[index] < preorder[s]) {
            index++;
        }
        
        root.left = build(preorder, s + 1, index - 1);
        root.right = build(preorder, index, e);
        return root;
    }
}
// @lc code=end