Medium
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 <= 3000inorder.length == preorder.length-3000 <= preorder[i], inorder[i] <= 3000preorder and inorder consist of unique values.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.using LeetCodeNet.Com_github_leetcode;
/**
* 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 {
private int j;
private Dictionary<int, int> map = new Dictionary<int, int>();
public int Get(int key) {
return map[key];
}
private TreeNode Answer(int[] preorder, int[] inorder, int start, int end) {
if (start > end || j > preorder.Length) {
return null;
}
int value = preorder[j++];
int index = Get(value);
TreeNode node = new TreeNode(value);
node.left = Answer(preorder, inorder, start, index - 1);
node.right = Answer(preorder, inorder, index + 1, end);
return node;
}
public TreeNode BuildTree(int[] preorder, int[] inorder) {
j = 0;
for (int i = 0; i < preorder.Length; i++) {
map.Add(inorder[i], i);
}
return Answer(preorder, inorder, 0, preorder.Length - 1);
}
}