Recursive Binary Tree Solution - Construct Binary Tree from Preorder and Inorder Traversal

The idea is that the 'first' element of Pre Order is always a root, and then count its position in In Order, then root.left = InOrder s + 1, s+count/s+count+1, end...

Recursive call.

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null || preorder.length != inorder.length || inorder.length == 0){
            return null;
        }
        if(preorder.length == 1){
            return new TreeNode(preorder[0]);
        }
       
        return buildTreeOps(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
    }
   
    private TreeNode buildTreeOps(int[] preorder, int[] inorder, int preS, int preE, int inS, int inE){
       
        if(preS > preE || inS > inE || preE > preorder.length || inE > inorder.length || (preE - preS) != (inE - inS)){
            return null;
        }
       
        if(preS == preE && inS == inE){
            return new TreeNode(inorder[inS]);
        }
        TreeNode node = new TreeNode(preorder[preS]);
        int count = 0;
        for(int i = 0; i <= (inE - inS); i++){
            if(inorder[i + inS] == preorder[preS]){
                count = i;
            }
        }
        node.left = buildTreeOps(preorder, inorder, preS + 1, preS + count, inS, inS + count - 1);
        node.right = buildTreeOps(preorder, inorder, preS + 1 + count, preE, inS + count + 1, inE);
        return node;
    }
}