判断一个数是否在二叉树中,如果存在返回真

#include<iostream>
#include<vector>
using namespace std;
//和链表的数据结构一样
struct TreeNode{
  
    int value;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x):value(x),left(NULL),right(NULL){}
};

bool  BST_search(TreeNode* node, int val){

    if(node->value  == val){
        return true;
    }
    if(node->value > val){
 
        if(node->left == NULL){
           return false;
        }else{
           BST_search(node->left, val);
        }
    }else{

        if(node->right == NULL){
            return false;
        }else{
            BST_search(node->right, val);
        }
    }

}


int main(){
    TreeNode a(8);
    TreeNode b(3);
    TreeNode c(10);
    TreeNode d(1);
    TreeNode e(6);
    TreeNode f(15);
    a.left =&b;
    a.right = &c;
    b.left = &d;
    b.right = &e;
    c.right = &f;
   
    for(int i = 0; i< 18; i++){
        if(BST_search(&a, i)){
            cout<<i<<"is in the BST"<<endl;
        }else{
            cout<<i<<"is not in the BST"<<endl;
        }
    
    }
      
    return 0;
}
二叉树的创建也可以通过类似于链表的方式创建,只是节点的链接关系不同

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转载自blog.csdn.net/liu_fei_er/article/details/80228055