Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than the node's key.
- Both the left and right subtrees must also be binary search trees.
Example 1:
2 / \ 1 3 Input: [2,1,3] Output: true
Example 2:
5 / \ 1 4 / \ 3 6 Input: [5,1,4,null,null,3,6] Output: false Explanation: The root node's value is 5 but its right child's value is 4.
// 06/09/2020
// Code golfed solution using pointers and short varaible names :-)
bool isValidBST(TreeNode* n, int* lo=nullptr, int* hi=nullptr){
return !n || ((!lo || n->val > *lo) && (!hi || n->val < *hi)
&& isValidBST(n->left, lo, &(n->val))
&& isValidBST(n->right, &(n->val), hi));
}
// 20/08/2020
// solution using optional instead of int64_t - branchless
bool isValidBST(TreeNode* root,
optional<int> minval = nullopt,
optional<int> maxval = nullopt){
return root == nullptr || (
(!minval || root->val > minval.value())
&& (!maxval || root->val < maxval.value())
&& isValidBST(root->left, minval, root->val)
&& isValidBST(root->right, root->val, maxval));
}
// 20/08/2020
// solution using optional instead of int64_t
bool isValidBST(TreeNode* root,
optional<int> minval = nullopt,
optional<int> maxval = nullopt){
if(root == nullptr) return true;
if((minval && root->val <= minval.value())
|| (maxval && root->val >= maxval.value())) return false;
return isValidBST(root->left, minval, root->val)
&& isValidBST(root->right, root->val, maxval);
}
// 20/08/2020
bool isValidBST(TreeNode* root,
int64_t minv=numeric_limits<int64_t>::min(),
int64_t maxv=numeric_limits<int64_t>::max()){
if(root == nullptr) return true;
if(root->val <= minv || root->val >= maxv) return false;
return isValidBST(root->left, minv, root->val)
&& isValidBST(root->right, root->val, maxv);
}