43: Construct Binary Tree from Preorder and Inorder Traversal

/************************************************************************/
            /*       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;
            }