Lintcode 前序遍历和中序遍历构造二叉树

public class 前序遍历和中序遍历构造二叉树 {

    /**
     * Definition of TreeNode:
     * public class TreeNode {
     *     public int val;
     *     public TreeNode left, right;
     *     public TreeNode(int val) {
     *         this.val = val;
     *         this.left = this.right = null;
     *     }
     * }
     */
     
    public static  class TreeNode {
              public int val;
             public TreeNode left, right;
              public TreeNode(int val) {
                 this.val = val;
                  this.left = this.right = null;
              }
     }
     
    public static class Solution {
        /**
         *@param preorder : A list of integers that preorder traversal of a tree
         *@param inorder : A list of integers that inorder traversal of a tree
         *@return : Root of a tree
         */
         public static TreeNode buildTree(int[] preorder, int[] inorder){
            if(preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0) 
                return null;
             return buildTree(preorder,inorder,0,preorder.length-1,0,inorder.length-1);
         }
         
        public static TreeNode buildTree(int[] preorder, int[] inorder,int pfrom ,int pto,int inbegin, int inend) {
            // write your code here
            if(pfrom > pto || inbegin > inend || inend >= inorder.length || pto>=preorder.length || pfrom < 0 || inbegin < 0) return null;
            System.out.println("pfrom" + pfrom + "pto" + pto + "inbeign" + inbegin + "inend" + inend);
              /**
              * 根节点元素肯定是preorder[pfrom];
              * preorder的第一个节点;
              */
             int rootdata = preorder[pfrom];
             TreeNode root = new TreeNode (rootdata);
             /**
              * 根据根节点元素查找到根节点在inorder中的位置;
              * 分成left和right两部分;
              */
             int rootpos = findRootPosition(inorder,inbegin,inend,rootdata);
             /**
              * 左边树的长度 rootpos - inbegin;
              * 右边树的长度 inend - rootpos;
              */
             int leftlen = rootpos - inbegin;//左边长度
             int rightlen =  inend - rootpos;//右边长度
             /**
              * 左边树的preorder从 pfrom + 1 到 pfrom+leftlen;
              * 左边树的inorder从inbegin到 rootpos - 1;
              * 
              * 右边树的preorder从 pto-rightlen到pto;注意(pto-rightlen+1)到pto
              * 右边树的inorder从rootpos+1 到inend;
              */
             TreeNode left =  buildTree(preorder,inorder,pfrom+1,pfrom+leftlen,inbegin,rootpos-1);
             TreeNode right = buildTree(preorder,inorder,pto-rightlen+1,pto,rootpos+1,inend);
             root.left = left;
             root.right = right;
             return root;
            
        }
        
        public static int findRootPosition(int inorder[],int begin, int  end ,int root){
            for(int i=begin;i<=end;i++)
                if(inorder[i] == root)
                    return i;
            return -1;//理论上都可以找到;找不到才返回-1
        }
        
        public static void main(String[] args) {
            int[] preorder = new int []{1,2};
            int[] inorder = new int[]{1,2};
            Solution.buildTree(preorder, inorder);
        }
    }

}