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

Note:
You may assume that duplicates do not exist in the tree.

见剑指offer重建二叉树

class Solution {     public TreeNode buildTree(int[] preorder, int[] inorder) {         return helper(0, 0, inorder.length - 1, preorder, inorder);     }          public TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {         if (preStart > preorder.length - 1 || inStart > inEnd)             return null;         TreeNode root = new TreeNode(preorder[preStart]);         int inIndex = 0;         for (int i = inStart; i <= inEnd; i++) {             if (inorder[i] == root.val)                 inIndex = i;         }         root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);         root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);         return root;     } }