【leetcode】Validate Binary Search Tree

Validate Binary Search Tree

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node's key.

The right subtree of a node contains only nodes with keys greater than the node's key.

Both the left and right subtrees must also be binary search trees.

confused what

"{1,#,2,3}"

means? > read more on how binary tree is serialized on OJ.

思路一：利用递归的思想，任何一个节点必须要在min和max范围内

min和max根据是左子树还是右子树来确定

如果是左子树：
node->left<node

如果是右子树：
node->right>node

递归条件：node->val在min和max范围内，node->left要在min,node->val范围内，node->right要在node->val,max范围内
node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val);

/**
* Definition for binary tree
* 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==NULL)
{
return true;
}

if(root->left==NULL&&root->right==NULL)
{
return true;
}

//注意如果测试用例含有INT_MAX则，必须采用long long int 才能避免出错
return testValid(root,(long long int)INT_MAX+1,(long long int)INT_MIN-1);
}

bool testValid(TreeNode *node,long long int max, long long int min)
{
if(node==NULL)
{
return true;
}

return node->val<max&&node->val>min&&testValid(node->left,node->val,min)&&testValid(node->right,max,node->val);

}
};

第二种方法，利用 二叉排序树 中序遍历 结果为递增序列的性质

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> ret_v;
bool isValidBST(TreeNode *root) {

if(root==NULL)
{
return true;
}

ret_v.clear();
inOrderTraversal(root);

//判断是否递增
for(int i=0;i<ret_v.size()-1;i++)
{
if(ret_v[i]>=ret_v[i+1])
{
return false;
}
}
return true;
}

//中序遍历，记录下数值
void inOrderTraversal(TreeNode *root)
{
if(root==NULL)
{
return;
}

inOrderTraversal(root->left);
ret_v.push_back(root->val);
inOrderTraversal(root->right);
}
};

第三种：直接在中序遍历中判断：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:

int ret;
bool flag;
bool isFirstNode;

bool isValidBST(TreeNode *root) {

flag=true;
//判断是不是第一个被访问的节点
isFirstNode=true;

inOrderTraversal(root);
return flag;
}

void inOrderTraversal(TreeNode *root)
{
if(root==NULL)
{
return;
}

if(!flag)
{
return;
}

inOrderTraversal(root->left);

//一旦发现不符合升序，则不是二叉排序树
if(!isFirstNode&&ret>=root->val)
{
flag=false;
return;
}

ret=root->val;
isFirstNode=false;

inOrderTraversal(root->right);
}

};