leetcode 之 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.
方法一：根据二叉查找树中序遍历是递增序列的特点，在进行中序遍历时，如果发现某个节点的值比它的先序节点要大，那么就是非法的。

class Solution {
public:
    bool isValidBST(TreeNode* root,TreeNode*& pre) {
    	if(!root)return true;
    	bool flag = isValidBST(root->left,pre);
    	if(!flag)return false;
    	if(pre && pre->val >= root->val)return false;
    	pre = root;
    	return isValidBST(root->right,pre);
    }
    
    bool isValidBST(TreeNode* root)
    {
    	TreeNode* pre = NULL;
    	return isValidBST(root,pre);
    }
};

方法二：根据二叉查找树的定义，每一节点都在左右子树值的中间，因此，每次递归遍历时，给定一个值的范围，该范围是当前子树值的两个边界，用该范围来判断合法性，该方法的效率比方法一高

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

class Solution {
public:
    bool isValidBST(TreeNode* root,int minValue,int maxValue)//左右子树的范围
    {
    	if(!root)return true;
    	if(minValue < root->val && root -> val < maxValue)
    	{
    		return isValidBST(root->left,minValue,root->val) && isValidBST(root->right,root->val,maxValue);
    	}
    	return false;
    }
    bool isValidBST(TreeNode* root)
    {
    	return isValidBST(root,INT_MIN,INT_MAX);
    }
};