Given preorder and inorder traversal of a tree, construct the binary tree.

Note:You may assume that duplicates do not exist in the tree.For example, given

preorder = [3,9,20,15,7]inorder = [9,3,15,20,7]Return the following binary tree:    3   / \  9  20    /  \   15   7

难度：Medium

题目： 给定二叉树的前序和中序遍历，构造二叉树，注意：二叉树结点值不重复。

思路：分治，

1. 在前序中找出根结点r，
2. 再在中序中找出r的位置，以该结点一分为二，继续执行1

Runtime: 8 ms, faster than 57.64% of Java online submissions for Construct Binary Tree from Preorder and Inorder Traversal.

Memory Usage: 37.5 MB, less than 49.23% of Java online submissions for Construct Binary Tree from Preorder and Inorder Traversal.

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        if (null == preorder || preorder.length < 1) {            return null;        }                return buildTree(preorder, 0, inorder, 0, inorder.length - 1);    }        private TreeNode buildTree(int[] preorder, int idx, int[] inorder, int s, int e) {        if (s > e) {            return null;        }        TreeNode root = new TreeNode(preorder[idx]);        int rIdx = s;        for (; rIdx <= e; rIdx++) {            if (inorder[rIdx] == preorder[idx]) {                break;            }        }        root.left = buildTree(preorder, idx + 1, inorder, s, rIdx - 1);        root.right = buildTree(preorder, idx + rIdx - s + 1, inorder, rIdx + 1, e);                return root;    }}