Careercup - Facebook面试题 - 5188884744896512

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 = ): 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 ;
 }