Licking Exercise - 58 Verifying Binary Search Trees

58 验证二叉搜索树

1.问题描述
给定一个二叉树,判断其是否是一个有效的二叉搜索树.

一个二叉搜索树具有如下特征：

节点的左子树只包含小于当前节点的数.

节点的右子树只包含大于当前节点的数.

所有左子树和右子树自身必须也是二叉搜索树.

示例 1:

输入:

2

/ \

1 3

输出: true

示例 2:

输入:

5

/ \

1 4

 / \

3   6

输出: false

解释: 输入为: [5,1,4,null,null,3,6].

 根节点的值为 5 ,但是其右子节点值为 4 .

可使用以下main函数：

#include

#include

#include

#include

#include

#include

using namespace std;

struct TreeNode

{

int val;

TreeNode *left;

TreeNode *right;

TreeNode() : val(0), left(NULL), right(NULL) {}

TreeNode(int x) : val(x), left(NULL), right(NULL) {}

TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

};

TreeNode* inputTree()

{

int n,count=0;

char item[100];

cin>>n;

if (n==0)

    return NULL;

cin>>item;

TreeNode* root = new TreeNode(atoi(item));

count++;

queue<TreeNode*> nodeQueue;

nodeQueue.push(root);

while (count<n)

{

    TreeNode* node = nodeQueue.front();

    nodeQueue.pop();

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int leftNumber = atoi(item);

        node->left = new TreeNode(leftNumber);

        nodeQueue.push(node->left);

    }

    if (count==n)

        break;

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int rightNumber = atoi(item);

        node->right = new TreeNode(rightNumber);

        nodeQueue.push(node->right);

    }

}

return root;

}

int main()

{

TreeNode* root;

root=inputTree();

bool res=Solution().isValidBST(root);

cout<<(res?"true":"false")<<endl;

}

2.输入说明
首先输入结点的数目n（注意,这里的结点包括题中的null空结点）

然后输入n个结点的数据,需要填充为空的结点,输入null.

The data of each node does not exceed32位intRepresentation range of type numbers.
3.输出说明
输出true或false
4.范例
输入
7
5 1 4 null null 3 6
输出
false
5.代码

#include <iostream>

#include <queue>

#include <climits>

#include<cstdlib>

#include <cstring>

#include<map>

#include<stack>
using namespace std;

struct TreeNode

{
    

	int val;

	TreeNode *left;

	TreeNode *right;

	TreeNode() : val(0), left(NULL), right(NULL) {
    }

	TreeNode(int x) : val(x), left(NULL), right(NULL) {
    }

	TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {
    }

};

TreeNode* inputTree()

{
    

	int n, count = 0;

	char item[100];

	cin >> n;

	if (n == 0)

		return NULL;

	cin >> item;

	TreeNode* root = new TreeNode(atoi(item));

	count++;

	queue<TreeNode*> nodeQueue;

	nodeQueue.push(root);

	while (count < n)

	{
    

		TreeNode* node = nodeQueue.front();

		nodeQueue.pop();

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int leftNumber = atoi(item);

			node->left = new TreeNode(leftNumber);

			nodeQueue.push(node->left);

		}

		if (count == n)

			break;

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int rightNumber = atoi(item);

			node->right = new TreeNode(rightNumber);

			nodeQueue.push(node->right);

		}

	}

	return root;

}

bool isValidBST(TreeNode *root)
{
    
	stack<TreeNode*> stack;//用栈实现中序遍历
	long long pre = (long long)INT_MIN - 1;//Considering that some sample values ​​will exceedINT
	while (!stack.empty() || root != NULL) {
    
		
		while (root != NULL) {
    
			stack.push(root);
			root = root->left;//Traverse to the bottom left element
		}
		root = stack.top();
		stack.pop();
		// 如果中序遍历得到的节点的值小于等于前一个 pre,说明不是二叉搜索树
		if (root->val <= pre) {
    
			return false;
		}
		pre = root->val;
		root = root->right;
	}
	return true;
}

int main()

{
    

	TreeNode* root;

	root = inputTree();

	bool res = isValidBST(root);

	cout << (res ? "true" : "false") << endl;

}