98. Validate Binary Search Tree

最新推荐文章于 2021-03-11 22:22:44 发布

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最新推荐文章于 2021-03-11 22:22:44 发布

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分类专栏： leetcode

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本文介绍了一种算法，用于判断给定的二叉树是否为有效的二叉搜索树。通过中序遍历的方式，确保左子树的节点值小于根节点，右子树的节点值大于根节点，且左右子树也必须是二叉搜索树。

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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:

Input:
2
/
1 3
Output: true

python:

class Solution(object):
    def isValidBST(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root==None:
            return True
        if root.left==None and root.right==None:
            return True
        self.list1=[]
        self.inOrder(root)
        for i in range(0,len(self.list1)-1):
            if self.list1[i+1]<=self.list1[i]:
                return False
        return True       
        
    def inOrder(self,root):
        if root==None:
            return
        self.inOrder(root.left)
        self.list1.append(root.val)
        self.inOrder(root.right)

c++:

class Solution {
public:
    
    bool isValidBST(TreeNode* root)
    {
        if(!root)
            return true;
        vector<int> vals;
        inOrder(root,&vals);
        for(int i=0;i<vals.size()-1;i++)
        {
            if(vals[i]>=vals[i+1])
                return false;
        }
        return true;     
        
    }
    void inOrder(TreeNode* root,vector<int> *vals)
    {
        if(!root)
            return;
        inOrder(root->left,vals);
        vals->push_back(root->val);
        inOrder(root->right,vals);
    }
};