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

time: O(nlogn) ~ O(n^2), space: O(height)

/**
 * 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 || preorder.length != inorder.length) {
            return null;
        }
        return buildTree(preorder, inorder, 0, 0, inorder.length - 1);
    }
    
    public TreeNode buildTree(int[] preorder, int[] inorder, int pre_start, int in_start, int in_end) {
        if(pre_start > preorder.length - 1 || in_start > in_end) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[pre_start]);
        int idx = 0;
        for(int i = in_start; i <= in_end; i++) {
            if(inorder[i] == preorder[pre_start]) {
                idx = i;
            }
        }
        
        root.left = buildTree(preorder, inorder, pre_start + 1, in_start, idx - 1);
        root.right = buildTree(preorder, inorder, pre_start + (idx - in_start + 1), idx + 1, in_end);
        
        return root;
    }
}