Leetcode Construct Binary Tree from preorder and inorder traversal

Given Preorder and inorder traversal of a tree, construct the binary tree.

Note:
Assume that duplicates does not exist in the tree.

Test instructions: Two forks the tree before, the middle order constructs two crosses the tree.

Idea: the classic topic.

/** * Definition for a binary tree node.  * public class TreeNode {*     int val, *     TreeNode left, *     TreeNode right; *     TreeNode (int x) {val = x;} *} */public class Solution {    TreeNode build (int[] preorder, int[] inorder, int l, int r, int len) {TreeNode root = null; if (Len < 1) return root;root = new TreeNode (preorder[l]), int index = 0;for (int i = 0; i < len; i++) if (inorder[r+ I] = = Preorder[l]) {index = I;break;} Root.left = Build (Preorder, Inorder, l+1, R, index); root.right = Build (Preorder, inorder, L+1+index, R+1+index, Len-1-inde x); return root;}    Public TreeNode Buildtree (int[] preorder, int[] inorder) {    int n = preorder.length;    Return build (preorder, inorder, 0, 0, n);}    }

Leetcode Construct Binary Tree from preorder and inorder traversal