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 keysless thanthe node's key.
• The right subtree of a node contains only nodes with keysgreater thanthe 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.

这道题就一个难点：要会设置节点的两边限制值。

吸收了LeetCode论坛上的建议，不使用INT_MAX 和INT_MIN，我这里使用了LLONG_MIN和LLONG_MAX。

因为如果是用INT_MIN，那么第一个左子树的值为INT_MIN的时候就会判断为假，其实为真。

虽然leetcode没有测试这个情况，不过健全的程序总是最好的。

class Solution {
public:
	bool isValidBST(TreeNode *root) 
	{
		return validBST(root);
	}

	//注意：别忘记了两边的boundary：leftMax和rightMax的设置
	/*
	I dont think it's a good idea to use int to represent the up and low bound of a TreeNode, INT_MIN and INT_MAX maybe used by TreeNode. We can use double or just the TreeNode itself.
	*/
	bool validBST(TreeNode *root, long long leftMax = LLONG_MIN, long long rightMax = LLONG_MAX) 
	{
		if (!root) return true;
		if (!root->left && !root->right) return true;
		if (root->left 
			&& (root->left->val >= root->val || root->left->val <= leftMax)) 
			return false;
		if (root->right 
			&& (root->right->val <= root->val || root->right->val >= rightMax )) 
			return false;

		return validBST(root->left, leftMax, root->val) 
			&& validBST(root->right, root->val, rightMax);
	}
};

//2014-2-15 update
	bool isValidBST(TreeNode *root) 
	{
		return isLegal(root);
	}
	bool isLegal(TreeNode *r, TreeNode *le = NULL, TreeNode *ri = NULL)
	{
		if (!r) return true;//||!r->left && !r->right不能增加这个条件
		if (le && le->val >= r->val || ri && ri->val <= r->val) return false;

		return isLegal(r->left, le, r) && isLegal(r->right, r, ri);
	}