106.BuildTree.cs

// 106. Construct Binary Tree from Inorder and Postorder Traversal
// Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary 
// tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

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

// Constraints:
// 1 <= inorder.length <= 3000
// postorder.length == inorder.length
// -3000 <= inorder[i], postorder[i] <= 3000
// inorder and postorder consist of unique values.
// Each value of postorder also appears in inorder.
// inorder is guaranteed to be the inorder traversal of the tree.
// postorder is guaranteed to be the postorder 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[] inorder, int[] postorder) {
        if(postorder is null || !postorder.Any() ||
           inorder is null || !inorder.Any()) 
        return null;

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