Given root of binary tree, determine if it's valid (left all < curr, right all > curr)
Inorder traversal & check if prev >= curr, recursive/iterative solutions
Time: O(n)
Space: O(n)
*/
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
bool isValidBST(TreeNode* root) {
return helper(root, LONG_MIN, LONG_MAX);
}
private:
bool helper(TreeNode* root, long left, long right){
if (!root)
return true;
if (root->val < right && root->val > left){
return helper(root->left, left, root->val) && helper(root->right, root->val, right);
}
return false;
}
};
class Solution {
public:
bool isValidBST(TreeNode* root) {
TreeNode* prev = NULL;
return inorder(root, prev);
}
private:
bool inorder(TreeNode* root, TreeNode*& prev) {
if (root == NULL) {
return true;
}
if (!inorder(root->left, prev)) {
return false;
}
if (prev != NULL && prev->val >= root->val) {
return false;
}
prev = root;
if (!inorder(root->right, prev)) {
return false;
}
return true;
}
};
*/