方法一（低效）：
最開始想的辦法是，先寫 求二叉樹中最大值的函數，和一個求二叉樹中最小值的函數，然后調用這兩個函數 寫出 驗證二叉搜索樹的函數（因為要左邊小于右邊嘛），結果 時間復雜度特別大，老是超時

/*** 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 isValidBST(TreeNode* root) {if(!root)return true;if(root&&root->left&&root->right){if(max_tree(root->left)<root->val&&min_tree(root->right)>root->val&&isValidBST(root->left)&&isValidBST(root->right))return true;else return false;}else{if(root->left){if(max_tree(root->left)<root->val&&isValidBST(root->left))return true;else return false;}else if(root->right){if(min_tree(root->right)>root->val&&isValidBST(root->right))return true;else return false;}else return true; }}int max_tree(TreeNode* root){if(root){//當根指針非空時int max = root->val;//最大值初始化為root的值if(root->left&&root->right){//當根指針的左右指針均非空時if(max_tree(root->left)>max|| max_tree(root->right)>max){//如果左右有一個大于maxif(max_tree(root->left)> max_tree(root->right))max = max_tree(root->left);else max = max_tree(root->right);} }else {//當左右指針至少有一個為空時if(root->left){if(max_tree(root->left)>max)max = max_tree(root->left);}else if(root->right){if(max_tree(root->right)>max)max = max_tree(root->right);}else; //當左右指針都為空時}return max; }else{return 0;} //當根指針為空時}int min_tree(TreeNode* root){if(root){//當根指針非空時int min = root->val;//最大值初始化為root的值if(root->left&&root->right){//當根指針的左右指針均非空時if(min_tree(root->left)<min|| min_tree(root->right)<min){//如果左右有一個if(min_tree(root->left)< min_tree(root->right))min = min_tree(root->left);else min = min_tree(root->right);} }else {//當左右指針至少有一個為空時if(root->left){if(min_tree(root->left)<min)min = min_tree(root->left);}else if(root->right){if(min_tree(root->right)<min)min = min_tree(root->right);}else; //當左右指針都為空時}return min; }else{return 0;} //當根指針為空時} };

方法二：利用中序遍歷，得到二叉樹的表，再查表看看是否是遞增的，如果是，則該二叉樹是二叉搜索樹，否則不是

#include<stdio.h> #include<limits.h> #include<string> #include<iostream> #include<vector>using namespace std; //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{ private:std::vector<int> vnums; public:bool isValidBST(TreeNode* root) {MOT(root);for(int ii=0;ii<vnums.size()-1;ii++){if(vnums[ii]>=vnums[ii+1])return false;}return true;}void MOT(TreeNode* root){if(root!=NULL){MOT(root->left);vnums.push_back(root->val);MOT(root->right);} }};int main(int argc,char* * argv) {struct TreeNode* root;root = (TreeNode*)malloc(sizeof(TreeNode));root->val = 4;root->left =(TreeNode*)malloc(sizeof(TreeNode)); root->right = (TreeNode*)malloc(sizeof(TreeNode));root->left->val =2; root->right->val = 6;root->left->left =(TreeNode*)malloc(sizeof(TreeNode)); root->left->right = (TreeNode*)malloc(sizeof(TreeNode));root->left->left ->val =1; root->left->right->val = 3;root->left->left->left=NULL; root->left->right->left=NULL;root->left->left->right=NULL; root->left->right->right=NULL;root->right->left =(TreeNode*)malloc(sizeof(TreeNode)); root->right->right = (TreeNode*)malloc(sizeof(TreeNode));root->right->left ->val =5; root->right->right->val = 7;root->right->left->left=NULL; root->right->right->left=NULL;root->right->left->right=NULL; root->right->right->right=NULL;solution s1;s1.isValidBST(root);return 0; }

總結

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