边工作边刷题：70天一遍leetcode: day 94-1

Largest BST Subtree

要点：

• http://articles.leetcode.com/largest-binary-search-tree-bst-in
• 这题重点是理解题意，还有道类似题 Largest Binary Search Tree (BST) in a Binary Tree – LeetCode( http://articles.leetcode.com/largest-binary-search-tree-bst-in_22
• 两题的区别：
  • 本题with all descendants：需要所有left/right子树都是bst，并且和root构成bst。而另一题是如果left or right不是bst，root还可以是（这里注意即使不要求所有descendants，结点仍需要连接的才构成bst）
  • 显然如果all descendants，需要检查到最底才能决定，所以bottom up方法。而另一题要从上到下扩展，所以top-down。
• return value和args：
  • 本题用bottom up的方法，所有min/max是return的，并且要return子树的判定状态和结点个数：-1表示子树不是bst，0个结点还是可以的。其他个数要比较min/max和root。
    • 通过检查-1/0来ignore返回的max/min：所以可以返回0/0作为max/min
  • 另一题return的个数是0或者实际个数，另外如果不能扩展了，要以这个子树为root重新计数

错误点：

• 不能因为left子树不符合就提前返回，仍然要遍历right子树来找到其他root对应的bst subtree

https://repl.it/Cb9Z/2

# 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:
#     10
#     / \
#   5  15
#   / \   \ 
#  1   8   7
# The Largest BST Subtree in this case is the highlighted one. 
# The return value is the subtree's size, which is 3.
# Hint:

# You can recursively use algorithm similar to 98. Validate Binary Search Tree at each node of the tree, which will result in O(nlogn) time complexity.
# Follow up:
# Can you figure out ways to solve it with O(n) time complexity?

# 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
        """
        def bfs(root):
            if not root:
                return 0,0,0
                
            isBST = True
            leftnodes, maxl, minl = bfs(root.left)
            if leftnodes==-1 or (leftnodes!=0 and root.val<maxl):
                isBST = False
            if leftnodes==0:
                minl = root.val
            
            rightnodes, maxr, minr = bfs(root.right)
            if rightnodes==-1 or (rightnodes!=0 and root.val>minr):
                isBST = False
            if rightnodes==0:
                maxr = root.val
            
            if isBST:
                self.maxLen = max(self.maxLen, leftnodes+rightnodes+1)
                return leftnodes+rightnodes+1, maxr, minl
            return -1, 0, 0
        
        self.maxLen = 0
        bfs(root)
        return self.maxLen