[leetcode] 98. Validate Binary Search Tree (medium)
2016-11-12 13:54
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题目:
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.
Example 1:
Binary tree
return true.
Example 2:
Binary tree
return false.
解题思路:
一开始使用类似前序遍历的方式进行判断,但是这样只能判断根节点的左右结点与他的大小关系,无法比较左右子树其他节点和根节点的大小关系,所以没有通过。
之后使用中序遍历的方式先把值存到数组里,再判断这个数组是不是递增的就可以了
代码:
class Solution {
public:
vector<int> v;
void inorder(TreeNode* root)
{
if(!root)
return;
inorder(root->left);
v.push_back(root->val);
inorder(root->right);
}
bool isValidBST(TreeNode* root) {
if(!root)
return 1;
inorder(root);
if(v.size()==1)
return 1;
for(int i=0;i<v.size()-1;i++)
if(v[i+1]<=v[i])
return 0;
return 1;
}
};
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.
Example 1:
2 / \ 1 3
Binary tree
[2,1,3],
return true.
Example 2:
1 / \ 2 3
Binary tree
[1,2,3],
return false.
解题思路:
一开始使用类似前序遍历的方式进行判断,但是这样只能判断根节点的左右结点与他的大小关系,无法比较左右子树其他节点和根节点的大小关系,所以没有通过。
之后使用中序遍历的方式先把值存到数组里,再判断这个数组是不是递增的就可以了
代码:
class Solution {
public:
vector<int> v;
void inorder(TreeNode* root)
{
if(!root)
return;
inorder(root->left);
v.push_back(root->val);
inorder(root->right);
}
bool isValidBST(TreeNode* root) {
if(!root)
return 1;
inorder(root);
if(v.size()==1)
return 1;
for(int i=0;i<v.size()-1;i++)
if(v[i+1]<=v[i])
return 0;
return 1;
}
};
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