Construct Binary Tree from Inorder and Postorder Traversal

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

Note:
You may assume that duplicates do not exist in the tree.

#include<stdio.h>#include<stdlib.h>typedef struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;}TreeNode;TreeNode *buildTreeCore(int *inorder, int inbegin, int inend, int *postorder, int postbegin, int postend) {    TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));    root->val = postorder[postend];    root->left = root->right = NULL;    if(inbegin == inend || postbegin == postend) return root;    int i;    for(i = inend; i >= inbegin && postorder[postend] != inorder[i]; i--) ;    if(i < inend) {        root->right = buildTreeCore(inorder, i + 1, inend, postorder, postbegin + i - inbegin, postend - 1);    }    if(i > inbegin) {        root->left = buildTreeCore(inorder, inbegin, i - 1, postorder, postbegin, postbegin + i - inbegin - 1);    }    return root;}TreeNode *buildTree(int *inorder, int *postorder, int n) {    if(inorder == NULL || postorder == NULL) return NULL;    TreeNode *root = buildTreeCore(inorder, 0, n - 1, postorder, 0, n - 1);}void Travel(TreeNode *root) {    if(root != NULL) {        printf("%d, ", root->val);        Travel(root->left);        Travel(root->right);    }}void main() {    int in[] = {1, 2, 3, 4, 5, 6};    int post[] = {1, 2, 4, 5, 6, 3};    TreeNode *root = buildTree(in, post, 6);    Travel(root);    printf("\n");}