105. 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.

public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
       if(preorder.length == 0) return null;
       return buildT(preorder, inorder ,0 ,preorder.length-1 ,0,inorder.length -1);
    }
    private TreeNode buildT(int[] preorder, int[] inorder ,int ps ,int pe ,int is,int ie){
       if(ps > pe || is > ie) return null;
       TreeNode root = new TreeNode(preorder[ps]);  //先序遍历第一个为父结点
       int i = is;
       for(; i <= ie;i++){                           //找出中序遍历的父节点，由此确定左右子树
           if(preorder[ps] == inorder[i]) break;
       }
       int left = i - is;
       root.left = buildT(preorder,inorder,ps+1,ps+left,is,i-1);  //DFS
       root.right = buildT(preorder,inorder,ps+left+1,pe,i+1,ie);
       return root;
    }
}