20230315 顺利通过
20230403 root == nullptr
&&
LONG_MIN, LONG_MAX
原题解
题目
约束
题解
解法一
class Solution {
public:
bool helper(TreeNode* root, long long lower, long long upper) {
if (root == nullptr) {
return true;
}
if (root -> val <= lower || root -> val >= upper) {
return false;
}
return helper(root -> left, lower, root -> val) && helper(root -> right, root -> val, upper);
}
bool isValidBST(TreeNode* root) {
return helper(root, LONG_MIN, LONG_MAX);
}
};
解法二
class Solution {
public:
bool isValidBST(TreeNode* root) {
stack<TreeNode*> stack;
long long inorder = (long long)INT_MIN - 1;
while (!stack.empty() || root != nullptr) {
while (root != nullptr) {
stack.push(root);
root = root -> left;
}
root = stack.top();
stack.pop();
// 如果中序遍历得到的节点的值小于等于前一个 inorder,说明不是二叉搜索树
if (root -> val <= inorder) {
return false;
}
inorder = root -> val;
root = root -> right;
}
return true;
}
};