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

Solution

/**
 * 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) {
        return helper(0, preorder.length - 1, 0, inorder.length - 1, preorder, inorder);
    }
    public TreeNode helper(int preStart, int preEnd, int inStart, int inEnd, int[] preorder, int[] inorder){
        if (preStart > preEnd || inStart > inEnd){
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = 0;
        for (int i = inStart; i <= inEnd; i++){
            if (inorder[i] == root.val){
                inIndex = i;
            }
        }
        root.left = helper(preStart + 1, preEnd, inStart, inIndex - 1, preorder, inorder);
        root.right = helper(preStart + 1 + inIndex - inStart, preEnd, inIndex + 1, inEnd, preorder, inorder);
        return root;
    }
}