Q4.5 Check if a binary search is BST

Q: Implement a function to check if a binary is a BST

A: DFS

根据平衡二叉树的定义，对于根节点root来说：

1、如果左孩子存在，那么root的值大于其左孩子的值

2、如果右孩子存在，那么root的值小于其右孩子的值

3、其左子树和右子树都是BST（递归判定）

满足上面的三个条件，该二叉树是BST，否则不是

#include <iostream>
#include <limits.h>
using namespace std;

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

bool checkBST(TreeNode *node, int lower, int upper, TreeNode *left, TreeNode *right) {
if (!node) {
return true;
}
return (node == left || node->val > lower) && (node == right || node->val < upper) && checkBST(node->left, lower, node->val, left, right) && checkBST(node->right, node->val, upper, left, right);
}

bool checkBST(TreeNode *root) {
if (!root) {
return true;
}
TreeNode *lefter = root; //最左节点
TreeNode *righter = root; //最右节点，为了兼容出现INT_MAX和INT_MIN的情况
while  (lefter->left) {
lefter = lefter->left;
}
while  (righter->right) {
righter = righter->right;
}
return checkBST(root, INT_MIN, INT_MAX, lefter, righter);
}

int main() {
int a[7] = {1,2,3,4,5,6,7};
TreeNode *root = new TreeNode(a[0]);
//	cout<<root->val<<endl;
int i = 1;
root->left = new TreeNode(a[i++]);
root->right = new TreeNode(a[i++]);
root->left->left = new TreeNode(a[i++]);
root->left->right = new TreeNode(a[i++]);
root->right->left = new TreeNode(a[i++]);
root->right->right = new TreeNode(a[i++]);
cout<<checkBST(root)<<endl;
return 0;
}