Construct Binary Tree from inorder and Postorder traversal

The original title is as follows;

Given Inorder and Postorder traversal of a tree, construct the binary tree.

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

Test instructions is simple: Spanning tree based on sequence traversal and sequential traversal

The idea is simple: according to the sequence of sequential traversal to determine the root node, in the middle sequence traversal to find the root node, the original sequence is divided into the left and right two parts, the recursive resolution of the left and right two parts can solve the problem.

The Java code is as follows:

public class Constructbinarytreefrominorderandpostordertraversal {/** * @Construct Binary Tree from Inorder and Postorder Traversal * link:https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/* Idea: Recursive */class treenode{int val; TreeNode left; TreeNode right;public TreeNode (int x) {val=x;}} Class Solution {public TreeNode buildtree (int[] inorder, int[] postorder) {if (inorder.length==0) {retur        n null;        } TreeNode node=createtree (inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);    return node; } public TreeNode Createtree (int[] inorder,int inbegin,int inend,int[] postorder,int postbegin,int postend) {if (    Inbegin>inend) {return null;    } int root=postorder[postend];    int index=0;    for (int i=inbegin;i<=inend;i++) {if (Root==inorder[i]) {index=i;    Break    }} int len=index-inbegin; TreeNode Left=createtree (inorder, Inbegin, Index-1, PostordeR, Postbegin, postbegin+len-1);    TreeNode Right=createtree (inorder, index+1, Inend, Postorder, Postbegin+len, postend-1);    TreeNode node=new TreeNode (root);    Node.left=left;    Node.right=right;    return node; }}}

Construct Binary Tree from inorder and Postorder traversal