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.

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null || preorder.length == 0 || inorder.length != preorder.length) return null;
        return buildTree(inorder, 0, preorder, 0, inorder.length);
    }
    
    public TreeNode buildTree(int[] inorder, int inStart, int[] preorder, int preStart, int length) {
        if (length == 0) return null;
        TreeNode root = new TreeNode(preorder[preStart]);
        int index = findIndex(inorder, root.val);
        root.left = buildTree(inorder, inStart, preorder, preStart + 1, index - inStart);
        root.right = buildTree(inorder, index + 1, preorder, preStart + (index - inStart) + 1, length - (index - inStart) - 1);
        return root;    
    }
    
    public int findIndex(int[] inorder, int val) {
        for (int i = 0; i < inorder.length; i++) {
            if (inorder[i] == val) {
                return i;
            }
        }
        return -1;
    }
}