leetCode(25):Validate Binary Search Tree 分类： leetCode 2015-06-23 13:00 154人阅读 评论(0) 收藏

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.

bool isValidBST(TreeNode* root)
{//判断一个二叉树是否是平衡二叉树
//首先平衡二叉树的任一结点的左孩子应该小于父亲结点
//任一结点的右孩子应该大于父亲结点
//任一结点的左子树的最大值应该小于它
//任一结点的右子树的最小值应该大于它
	if(NULL==root)
		return true;
	
	TreeNode* behind=root->right;
	while(behind && behind->left)
		behind=behind->left;
	TreeNode* ahead=root->left;
	while(ahead && ahead->right)
		ahead=ahead->right;
	
	//左右子树均不为空的情况
	if(behind && ahead)
	{
		if(behind->val > root->val && ahead->val < root->val)
		{
			if(root->left->val < root->val && root->right->val > root->val)
				return isValidBST(root->left) && isValidBST(root->right);
			else
				return false;
		}
		else
			return false;
	}
	
	//右子树不为空的情况
	if(behind)
	{
		if(behind->val > root->val && root->right->val > root->val)
			return isValidBST(root->right);
		else
			return false;
	}
	
	//左子树不为空的情况
	if(ahead)
	{
		if(ahead->val < root->val && root->left->val < root->val)
			return isValidBST(root->left);
		else
			return false;
	}
	
	//左右子树均为空的情况
	return true;
}