Post-order traversal sequence of the Binary Search Tree

Enter an integer array to determine whether the array is the result of a sorted traversal of a binary search tree. If Yes, Yes is output; otherwise, No is output. Assume that any two numbers in the input array are different. Input: each test case contains two rows. The first row is an integer of n (1 <= n <= 10000), indicating the length of the array. The second row contains n integers, indicating the array. The value range of the number in the array is [0,]. Output: corresponds to each test case. If the input array is a result of a forward traversal of a binary search tree, Yes is output; otherwise, No is output. Sample input: 75 7 6 9 11 10 847 4 6 5 sample output: YesNo code AC: idea: the tree needs to be traversed. The binary tree is characterized by lchild. key <root. key <rchild. then we use the sub-governance idea to separate left and right sub-trees by using the above features. The first sub-tree is greater than root. after the key (right side) is the sub-tree, If we encounter less than root when traversing the sub-tree. in the case of key, it is the wrong sequence .... [Cpp] // use sub-governance! # Include <stdio. h> # include <stdlib. h> long int * num; int fun (long int low, long int high) {long int I, l, tmp_idx, flag = 1; if (low> = high) {return 1 ;}for (I = low; I