NO.106 sequence from the sequence postorder binary tree structure

Traversing the binary tree is configured in accordance with the sequence preorder a tree.

Note:
You can assume that the tree does not duplicate elements.

For example, given

Preorder inorder = [9,3,15,20,7]
Sequence after postorder traversal = [9,15,7,20,3]

Returns the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */


struct TreeNode* buildTree(int* inorder, int inorderSize, int* postorder, int postorderSize){
    if(postorderSize==0||postorderSize!=inorderSize)return NULL;
    struct TreeNode* ret=(struct TreeNode *)malloc(sizeof(struct TreeNode));
    ret->val=postorder[postorderSize-1];
    int left_len=0;
    for(int i=0;i<inorderSize;i++)
    {
        if(inorder[i]==postorder[postorderSize-1])break;
        left_len++;
    }
    assert(left_len<inorderSize);
    ret->left=buildTree(inorder,left_len,postorder,left_len);
    ret->right=buildTree(inorder+1+left_len,inorderSize-1-left_len,postorder+left_len,inorderSize-1-left_len);
    return ret;
}

When execution: 28 ms, beat the 93.88% of all users in C submission

Memory consumption: 13.1 MB, defeated 100.00% of all users in C submission