标签：init root null max || 最小值 validate 思路 bsp
思路1：对树进行深度优先遍历，遍历到某个节点是判断该节点的左子树是否为二叉搜索树，右子树是否为二叉搜索树，左子树的最大值是否小于当前节点的值，右子树的最小值是否大于当前节点的值。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isValidBST(TreeNode* root) { 13 if(root == NULL) return true; 14 int maxv, minv; 15 return dfs(root, maxv, minv); 16 } 17 18 bool dfs(TreeNode* root, int &maxv, int &minv){ 19 maxv = minv = root->val; 20 if(root->left){ 21 int nowMaxv, nowMinv; 22 if(!dfs(root->left, nowMaxv, nowMinv)) 23 return false; 24 if(nowMaxv >= root->val) 25 return false; 26 maxv = max(maxv, nowMaxv); 27 minv = min(minv, nowMinv); 28 } 29 if(root->right){ 30 int nowMaxv, nowMinv; 31 if (!dfs(root->right, nowMaxv, nowMinv)) 32 return false; 33 if (nowMinv <= root->val) 34 return false; 35 maxv = max(maxv, nowMaxv); 36 minv = min(minv, nowMinv); 37 } 38 return true; 39 } 40 };
思路2：对树进行中序遍历，遍历的过程中判断当前值是否大于前一个值。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isValidBST(TreeNode* root) { 13 stack<TreeNode*> st; 14 int pre_val; 15 bool first_flag = true; 16 while(!st.empty() || root!=NULL){ 17 while(root!=NULL){ 18 st.push(root); 19 root = root->left; 20 } 21 root = st.top(); 22 st.pop(); 23 if(first_flag) 24 first_flag = false; 25 else if(root->val <= pre_val) 26 return false; 27 pre_val = root->val; 28 root = root->right; 29 } 30 return true; 31 } 32 };
标签：init root null max || 最小值 validate 思路 bsp
原文地址：https://www.cnblogs.com/sclczk/p/11227142.html
