這題要多想，首相二叉平衡樹的判斷一定要從全局來看，分別大于小于某個數(shù)！！！，第二點，我被最大最小值卡了，所以就用了long long int然后改最大最小值，2刷刷好看一點的代碼，思路要更加清晰
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: bool mark; void findit(TreeNode* root, long long int minn, long long int maxx){ if(root == NULL) return ; if(mark == false) return ; if(!(root -> val < maxx && root -> val > minn)){ mark = false; return ; } if(root -> left != NULL){ if(root -> left -> val >= root -> val){ mark = false; return ; } findit(root -> left, minn, root -> val); } if(root -> right != NULL){ if(root -> right -> val <= root -> val){ mark = false; return ; } findit(root -> right, root -> val, maxx); } return ; } bool isValidBST(TreeNode* root) { mark = true; if(root == NULL) return true; findit(root, -2147483649, 2147483648); return mark; }};新聞熱點
疑難解答
