Largest BST Subtree

Given a binary tree, find the largest subtree which is a Binary Search Tree (BST), where largest means subtree with largest number of nodes in it.

Note:
A subtree must include all of its descendants.
Here's an example:

The Largest BST Subtree in this case is the highlighted one.
The return value is the subtree's size, which is 3.

Follow up:
Can you figure out ways to solve it with O(n) time complexity?

在二叉树中查找为最大BST的子树, 开始时觉得这题和剑指offer上Longest increasing subarray是一类题目.只是从数组换成了动态的中序遍历. 后来发现这种情况无法检验是否是一整颗子树.所以需要换一种策略来做.也就是devide and conquer,对于每个节点,考虑其左右子树是否是BST, 如果都是是BST,则再检验加上这个节点之后组成的子树是否是BST. 即leftMax < root.val && rightMin > root.val. 之后计算当前子树的大小,需要利用左右子树的大小,所以这题每次遍历需要返回多个变量: size, min, max. 当然也需要返回当前子树是否是BST的一个状态信息,可以加上一个flag变量, 或者我们也可以利用Balanced Binary Tree

中的做法,大于等于0的确实表示高度,-1表示不平衡.所以这里我们可以用size = -1表示子树非BST.每个节点处理一次,时间复杂度为O(n),空间复杂度为栈高度O(logn),代码如下:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def largestBSTSubtree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        res = [0]
        Min, Max, size = self.helper(root, res)
        return res[0]
    def helper(self, root, res):
        if not root:
            return sys.maxint, -sys.maxint -1, 0 #注意节点不存在时,为了方便比较,Min定义为最小值,Max定义为最大值.或者inf
            #return 0, float('inf'), float('-inf')
        leftMin, leftMax, leftSize = self.helper(root.left, res)
        rightMin, rightMax, rightSize = self.helper(root.right, res)
        if leftSize == -1 or rightSize == -1 or leftMax > root.val or rightMin < root.val:
            return 0, 0, -1
        else:
            size = leftSize + rightSize + 1
            res[0] = max(res[0], size)
            return min(leftMin, root.val), max(root.val, rightMax), size #除去sys.maxint和sys.maxint的影响.

原始未优化代码:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def largestBSTSubtree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        res = [0]
        Min, Max, flag, size = self.helper(root,res)
        return res[0]
    def helper(self, root, res):
        if not root.left and not root.right:
            res[0] = max(res[0],1)
            return root.val, root.val, True, 1
        Min = Max = root.val
        size = -1
        flag = True
        leftSize = rightSize = 0
        if root.left:
            leftMin, leftMax, leftFlag, leftSize = self.helper(root.left, res)
            if not leftFlag or leftMax > root.val:
                flag = False
            else:
                Min = leftMin
        if root.right:
            rightMin, rightMax, rightFlag, rightSize = self.helper(root.right, res)
            if not rightFlag or rightMin < root.val:
                flag = False
            else:
                Max = rightMax
        if not flag:
            return -1, -1, False, size
        else:
            size = 1 + leftSize + rightSize
            res[0] = max(res[0], size)
            return Min, Max, True, size

转载于:https://www.cnblogs.com/sherylwang/p/5666115.html

猜你喜欢