Leetcode Validate Binary Search Tree problem solution YASH PAL, 31 July 2024 In this Leetcode Validate Binary Search Tree problem solution we have Given the root of a binary tree, determine if it is a valid binary search tree (BST). A valid 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. Problem solution in Python. class Solution: def isValidBST(self, root: TreeNode) -> bool: def inorder(root): if root.left: inorder(root.left) res.append(root.val) if root.right: inorder(root.right) res=[] if root: inorder(root) for i in range(1,len(res)): if res[i-1]>=res[i]: return False return True Problem solution in Java. List<Integer> sol = new ArrayList<>(); public boolean isValidBST(TreeNode root) { dfs(root); for(int i=1; i<sol.size(); i++) if(sol.get(i-1)>=sol.get(i)) return false; return true; } public void dfs(TreeNode root) { if(root==null) return; dfs(root.left); sol.add(root.val); dfs(root.right); } Problem solution in C++. class Solution { public: bool isValidBST(TreeNode* root) { return helper(root,NULL,NULL); } bool helper(TreeNode* root, TreeNode* min, TreeNode* max) { if(root==NULL) return true; else if(min!=NULL && root->val<=min->val || max!=NULL && root->val>=max->val) return false; else if( helper(root->left, min, root) && helper(root->right, root, max)) return true; return false; } }; Problem solution in C. bool check(struct TreeNode* root, int* min, int* max) { if (root == NULL) { return true; } if ((min != NULL && root->val <= *min) || (max != NULL && root->val >= *max)) { return false; } return check(root->left, min, &root->val) && check(root->right, &root->val, max); } bool isValidBST(struct TreeNode* root){ return check(root, NULL, NULL); } coding problems