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

1 /** 2  * Definition for a binary tree node. 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 class Solution {11     public TreeNode buildTree(int[] preorder, int[] inorder) {12         if (preorder == null || inorder == null || preorder.length == 0 || inorder.length != preorder.length) return null;13         return buildTree(inorder, 0, preorder, 0, inorder.length);14     }15     16     public TreeNode buildTree(int[] inorder, int inStart, int[] preorder, int preStart, int length) {17         if (length == 0) return null;18         TreeNode root = new TreeNode(preorder[preStart]);19         int index = findIndex(inorder, root.val);20         root.left = buildTree(inorder, inStart, preorder, preStart + 1, index - inStart);21         root.right = buildTree(inorder, index + 1, preorder, preStart + (index - inStart) + 1, length - (index - inStart) - 1);22         return root;    23     }24     25     public int findIndex(int[] inorder, int val) {26         for (int i = 0; i < inorder.length; i++) {27             if (inorder[i] == val) {28                 return i;29             }30         }31         return -1;32     }33 }