LeetCode --- 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.

解题思路：按先序序列和中序序列构造二叉树，先序序列中第一个值为根的值。递归遍历如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode constructBTree(int [] preorder,int pstart,int pend,int[] inorder,int istart,int iend){
     int index=-1;
     TreeNode node=new TreeNode(preorder[pstart]);
     for(int i=istart;i<=iend;i++){
      if(inorder[i]==preorder[pstart]){
      index=i;
      break;
      } 
     
     }
    
     if(istart<index){
     int len=index-istart;
     node.left=constructBTree(preorder,pstart+1,pstart+len,inorder,istart,index-1);
     
     }
     
     if(index<iend){
     int len=iend-index;
     node.right=constructBTree(preorder,pend-len+1,pend,inorder,index+1,iend);
     }
         
     return node;
    }

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root=null;
        int len=preorder.length;
        if(len==0||preorder==null){
        return null;
        }  
       root=constructBTree(preorder,0,len-1,inorder,0,len-1);  
        return root;
    }
}