Careercup

2014-05-02 07:18

题目链接

原题：

boolean isBST(const Node* node) { 
// return true iff the tree with root 'node' is a binary search tree. 
// 'node' is guaranteed to be a binary tree. 
} 

  n
 / 
a   b
     
      c

题目：检查一棵二叉树是否为二叉搜索树。

解法：二叉搜索树，就是每个节点的左边全都小于它，右边全都大于它。如果真的对于每个节点都全部检查左右边的每个节点，就做了很多重复劳动。只需要判断每个节点的左子树最靠右，和右子树最靠左的节点是否小于和大于它即可。递归过程中传递引用可以随时更新两个需要检查的值。请看代码。

代码：

// http://www.careercup.com/question?id=5632735657852928
#include <climits>
#include <iostream>
#include <sstream>
#include <string>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int _val = 0): val(_val), left(nullptr),right(nullptr) {};
};

class Solution {
public:
    bool isBST(TreeNode *root) {
        if (root == nullptr) {
            return false;
        }
        
        max_val = INT_MIN;
        res = true;
        first_node = true;
        isBSTRecursive(root);
        
        return res;
    };
private:
    int max_val;
    bool res;
    bool first_node;
    
    void isBSTRecursive(TreeNode *root) {
        if (!res) {
            return;
        }
        
        // root is guaranteed to be not nullptr.
        if (root->left) {
            isBSTRecursive(root->left);
        }
        if (first_node || root->val > max_val) {
            first_node = false;
            max_val = root->val;
        } else {
            res = false;
            return;
        }
        if (root->right) {
            isBSTRecursive(root->right);
        }
    };
};

void construcTree(TreeNode *&root)
{
    int val;
    stringstream sio;
    string s;
    
    if (cin >> s && s != "#") {
        sio << s;
        sio >> val;
        root = new TreeNode(val);
        construcTree(root->left);
        construcTree(root->right);
    } else {
        root = nullptr;
    }
}

void deleteTree(TreeNode *&root)
{
    if (root == nullptr) {
        return;
    }
    deleteTree(root->left);
    deleteTree(root->right);
    delete root;
    root = nullptr;
}

int main()
{
    TreeNode *root;
    Solution sol;
    
    while (true) {
        construcTree(root);
        if (root == nullptr) {
            break;
        }
        
        cout << (sol.isBST(root) ? "Valid BST" : "Invalid BST") << endl;
        
        deleteTree(root);
    }
    
    return 0;
}