Validate Binary Search Tree
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.
分析:采用递归的方法,先判断左子树和右子树是不是BST,如果不是返回false,如果左子树和右子树均为BST,那么再比较左子树最大值、根的值、右子树最小值的关系,如果左子树最大值<根值<右子树最小值,则返回true.代码如下:
class Solution { public: bool isValidBST(TreeNode *root) { if(!root) return true; if((!root->left || root->val > tree_max(root->left)) && (!root->right || root->val < tree_min(root->right))) return isValidBST(root->left) && isValidBST(root->right); return false; } int tree_max(TreeNode *root){ if(!root) return INT_MIN; int result = root->val; while(root->right) { root = root->right; result = root->val; } return result; } int tree_min(TreeNode *root){ if(!root) return INT_MAX; int result = root->val;; while(root->left){ root = root->left; result = root->val; } return result; } };
上面代码有点繁琐,因为有两个函数tree_max和tree_min来计算树的最大值和最小值,以下代码通过维护根值的上限和下限代替最大值最小值,代码简单明了:
1 class Solution { 2 public: 3 bool isValidBST(TreeNode *root) { 4 return isValidBST(root, LLONG_MIN, LLONG_MAX); 5 } 6 bool isValidBST(TreeNode* root, long long lower, long long upper) { 7 if (root == nullptr) return true; 8 return root->val > lower && root->val < upper 9 && isValidBST(root->left, lower, root->val) 10 && isValidBST(root->right, root->val, upper); 11 } 12 };
上面上限和下限用long long类型的原因是结点值可能是INT_MIN和INT_MAX,为防止出错,所以用long long类型。