[LeetCode] Validate Binary Search Tree
方法一:若是bst,中序便利后一定是有序,递增的。可以中序遍历后查看是否递增来判断
1 class Solution { 2 public: 3 bool isValidBST(TreeNode *root) { 4 5 if(root == NULL) 6 return true; 7 vector<int> result; 8 stack<TreeNode*> st; 9 10 TreeNode* p = root; 11 12 // inorder traverse 13 while(p != NULL || st.size() != 0) 14 { 15 while(p != NULL) 16 { 17 st.push(p); 18 p = p->left; 19 } 20 21 if(!st.empty()) 22 { 23 p = st.top(); 24 st.pop(); 25 result.push_back(p->val); 26 p = p->right; 27 } 28 } 29 30 //check if it is ascend 31 for(int i = 0; i < result.size()-1; i++) 32 { 33 if(result[i] < result[i+1]) 34 ; 35 else 36 return false; 37 } 38 return true; 39 40 } 41 };
方法二:使用递归,判断当前节点的值,是否在上边界和下边界之间,对于根节点,就是没有边界。然后递归判断
1 class Solution { 2 public: 3 bool isValidBST(TreeNode *root) { 4 return isValidBST(root, INT_MIN, INT_MAX); 5 } 6 bool isValidBST(TreeNode *root, int min, int max) 7 { 8 if(root == NULL) return true; 9 10 if( (root->val < max) && (root->val > min) && 11 isValidBST(root->left, min, root->val) && 12 isValidBST(root->right, root->val, max) 13 ) 14 return true; 15 else 16 return false; 17 } 18 };
方法3: 方法二没有Ac,由于两个INT_MIN出现时,所以考虑用long long 代替int即可
class Solution { public: bool isValidBST(TreeNode *root) { return isValidBST(root, (long long)INT_MIN - 1, (long long)INT_MAX + 1); } bool isValidBST(TreeNode *root, long long min, long long max) { if(root == NULL) return true; if( (root->val < max ) && (root->val > min) && isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max) ) return true; else return false; } };