1008. 先序遍历构造二叉树

解法一

根据二叉搜索树的性质找到左右子树结点在数组中的范围，递归建树。

/**
 * 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 {
    public TreeNode bstFromPreorder(int[] preorder) {
        return build(preorder, 0, preorder.length);
    }
    private TreeNode build(int[] nums, int begin, int end) {
        if (begin >= end) {
            return null;
        }
        TreeNode root = new TreeNode(nums[begin]);
        int mid = begin + 1;
        while ((mid < end) && (nums[mid] < nums[begin])) {
            ++mid;
        }
        root.left = build(nums, begin + 1, mid);
        root.right = build(nums, mid, end);
        return root;
    }
}