Using preorder traversal binary tree structure, and preorder

Traversing the binary tree is configured in accordance with the sequence preorder traversal of a tree. 

Note: 
 You can assume that the tree does not duplicate elements. 
 
 For example, given 
 
 a preorder traversal preorder = [ 3 , . 9 , 20 is , 15 , . 7 ] 
 preorder InOrder = [ . 9 , 3 , 15 , 20 is , . 7 ] 
 returns the following binary tree: 

     3 
    / \
    . 9   20 is 
     /   \
     15    . 7

Idea: using the idea of ​​divide and conquer to solve the problem

Specific problem-solving steps:

　　1. The preorder traversal, we can know that the first element is the root of the given array pre [0], then we can find a value equal to the pre preorder [0] position, the first half of the location part is the left subtree, the right half is the right subtree,

　　2. Repeat 1 until traversed

Codes are as follows:

public class Solution {

    public int preIndex = 0;

    //查找pre[pri]的位置
    public int find(int[] inorder, int inbegin,int inend,int val){
        for(int i = inbegin;i<=inend;i++){
            if(inorder[i] == val) return i;
        }
        return -1;
    }

    public TreeNode buildTreeChild(int[] preorder, int[] inorder, intinbegin, int inend) {
         if (inbegin> inend) return  null ; 
        TreeNode root = new TreeNode (preorder [preIndex]);
        int index = find (ordered, inbegin, inend, preorder [preIndex]);
        if (index == - 1 ) return  null ; 
        preIndex ++ ; 
        root.left = buildTreeChild (preorder, inorder, inbegin, index- 1 ); 
        root.right = buildTreeChild (preorder, inorder, index + 1 , inend);
        return root; 
    }

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