/************************************************************************/

            /*       43:  Construct Binary Tree from Preorder and Inorder Traversal                            */

            /************************************************************************/

            /*

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

             *

             * Note:

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

             * */

            

            /*** 递归做法*********************/

            /*

             * 关键的地方在于：

             *

             * 在前序遍历的序列里，同时对左子树和右子树进行递归得到一个节点的左右节点

             * */

public TreeNode buildTreeByPre_In(int[] preorder, int[] inorder)            {                 return helper(0, 0, inorder.length - 1, preorder, inorder);            }                        private  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; // Index of current root in inorder                for (int i = inStart; i <= inEnd; i++) {                    if (inorder[i] == root.val) {                        inIndex = i;                        break;                    }                }                root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);                root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);                return root;            }