105.BuildTree.cs

// 105. Construct Binary Tree from Preorder and Inorder Traversal
// Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary 
// tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

// Example 1:
// Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
// Output: [3,9,20,null,null,15,7]
// Example 2:
// Input: preorder = [-1], inorder = [-1]
// Output: [-1]

// Constraints:
// 1 <= preorder.length <= 3000
// inorder.length == preorder.length
// -3000 <= preorder[i], inorder[i] <= 3000
// preorder and inorder consist of unique values.
// Each value of inorder also appears in preorder.
// preorder is guaranteed to be the preorder traversal of the tree.
// inorder is guaranteed to be the inorder traversal of the tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public TreeNode BuildTree(int[] preorder, int[] inorder) {
        if(preorder is null || !preorder.Any() ||
           inorder is null || !inorder.Any()) 
        return null;

        TreeNode root = new TreeNode(preorder[0]);
        int middle = Array.IndexOf(inorder, preorder[0]);
        root.left = BuildTree(preorder[1..(middle+1)], inorder[..middle]);
        root.right = BuildTree(preorder[(middle+1)..], inorder[(middle+1)..]);
        
        return root;
    }
}