边工作边刷题：70天一遍leetcode: day 76

Count Univalue Subtrees

要点：检测条件比较有意思：因为可能的情况比较多，只要违反了任意一条就return False，所以可以只考虑False的情况，最后return True。

https://repl.it/CoqQ

• 错误点：这题类似Largest BST Subtree（with all descendants，bottom up方法，or post order），左子树不符合只能invalidate root，但仍然要recurse到右子树，所以不能提前返回，而是要在右子树访问完以后返回。

post-order iteration: 注意必须要用map记录，因为post-order没法从栈中获得当前结点左/右子树的情况，所以只能用map记录

https://repl.it/CopV

# 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 countUnivalSubtrees(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def helper(root):
            isUnival = True
            if root.left and (not helper(root.left) or root.val!=root.left.val):
                isUnival = False

            if root.right and (not helper(root.right) or root.val!=root.right.val):
                isUnival = False

            if not isUnival:
                return False
            #print root.val
            self.count+=1
            return True

        self.count=0
        if not root: return 0
        helper(root)
        return self.count

# Given a binary tree, count the number of uni-value subtrees.

# A Uni-value subtree means all nodes of the subtree have the same value.

# For example:
# Given binary tree,
#               5
#              / \
#             1   5
#           / \   \
#           5   5   5
# return 4.

# Hide Tags Tree

# 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 countUnivalSubtrees(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root: return 0
        stk = [root]
        umap = {}
        pre = None
        count = 0
        while stk:
            cur = stk[-1]
            if pre==None or pre.left==cur or pre.right==cur:
                if cur.left:
                    stk.append(cur.left)
                elif cur.right:
                    stk.append(cur.right)
            elif cur.left==pre:
                if cur.right:
                    stk.append(cur.right)
            else:
                stk.pop()
                if cur.left and (not umap[cur.left] or cur.left.val!=cur.val):
                    pre = cur
                    umap[cur]=False
                    continue

                if cur.right and (not umap[cur.right] or cur.right.val!=cur.val):
                    pre = cur
                    umap[cur]=False
                    continue
                umap[cur]=True
                count+=1
            pre = cur
        return count