力扣算法题—098验证二叉搜索树
给定一个二叉树,判断其是否是一个有效的二叉搜索树。
假设一个二叉搜索树具有如下特征:
- 节点的左子树只包含小于当前节点的数。
- 节点的右子树只包含大于当前节点的数。
- 所有左子树和右子树自身必须也是二叉搜索树。
示例 1:
输入: 2 / \ 1 3 输出: true
示例 2:
输入: 5 / \ 1 4 / \ 3 6 输出: false 解释: 输入为: [5,1,4,null,null,3,6]。 根节点的值为 5 ,但是其右子节点值为 4 。
1 #include "_000库函数.h" 2 3 #include <queue> 4 5 struct TreeNode { 6 int val; 7 TreeNode *left; 8 TreeNode *right; 9 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 10 }; 11 12 //解法一: 13 class Solution { 14 public: 15 bool isValidBST(TreeNode* root) { 16 return helper(root); 17 } 18 19 bool helper(TreeNode* root) { 20 if (root->left == NULL && root->right == NULL) 21 return true; 22 if (root->left == NULL || root->right == NULL) 23 return false; 24 if (root->left->val >= root->val || root->right->val <= root->val) 25 return false; 26 return helper(root->left); 27 helper(root->right); 28 } 29 }; 30 31 32 //解法二: 33 // Recursion without inorder traversal 34 class Solution { 35 public: 36 bool isValidBST(TreeNode* root) { 37 return isValidBST(root, LONG_MIN, LONG_MAX); 38 } 39 bool isValidBST(TreeNode* root, long mn, long mx) { 40 if (!root) return true; 41 if (root->val <= mn || root->val >= mx) return false; 42 return isValidBST(root->left, mn, root->val) && isValidBST(root->right, root->val, mx); 43 } 44 }; 45 46 47 //解法三:中序遍历 48 // Recursion 49 class Solution { 50 public: 51 bool isValidBST(TreeNode* root) { 52 if (!root) return true; 53 vector<int> vals; 54 inorder(root, vals); 55 for (int i = 0; i < vals.size() - 1; ++i) { 56 if (vals[i] >= vals[i + 1]) return false; 57 } 58 return true; 59 } 60 void inorder(TreeNode* root, vector<int>& vals) { 61 if (!root) return; 62 inorder(root->left, vals); 63 vals.push_back(root->val); 64 inorder(root->right, vals); 65 } 66 }; 67 68 //解法4: 69 class Solution { 70 public: 71 bool isValidBST(TreeNode* root) { 72 if (!root) 73 return true; 74 queue<TreeNode* >q; 75 q.push(root); 76 helper(q, root); 77 while (!q.empty()) { 78 int a = q.front()->val; 79 q.pop(); 80 int b = q.front()->val; 81 if (a >= b) 82 return false; 83 } 84 } 85 void helper(queue<TreeNode* >&q, TreeNode* root) { 86 if (!root) 87 return; 88 helper(q, root->left); 89 q.push(root); 90 helper(q, root->right); 91 } 92 }; 93 94 95 TreeNode* CreateBTree(const vector<int> &v) 96 { 97 if (v.size() == 0) 98 return nullptr; 99 TreeNode* root = new TreeNode(v[0]); 100 auto cursor = root; 101 queue<TreeNode*> que; 102 que.push(cursor); 103 for (size_t i = 1; i < v.size(); i++) 104 { 105 if (que.empty()) 106 break; 107 cursor = que.front(); 108 if (cursor == NULL) 109 break; 110 que.pop(); 111 112 if (v[i] != 0) 113 cursor->left = new TreeNode(v[i]); 114 if (v[++i] != 0) 115 cursor->right = new TreeNode(v[i]); 116 117 que.push(cursor->left); 118 que.push(cursor->right); 119 } 120 return root; 121 } 122 123 124 125 126 void T098() { 127 Solution s; 128 vector<int>v = { 6,5,7,0,0,6,8}; 129 TreeNode *root; 130 131 root = CreateBTree(v); 132 cout << s.isValidBST(root) << endl; 133 134 v = { 2,1,3 }; 135 root = CreateBTree(v); 136 cout << s.isValidBST(root) << endl; 137 }