# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution1(object):
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
return self.countNodes(root.left)+self.countNodes(root.right)+1
# https: // segmentfault.com / a / 1190000011040217
# https://blog.csdn.net/qq_17550379/article/details/82052746
# 如果右子树的高度等于整棵二叉树的高度-1的话，那么左子树一定是一棵满二叉树，
# 这个时候我们就很容易的计算出总结点数nodes=2**(h)-1 + 1 +右子树节点数（这里的+1表示root节点）。
# 如果右子树的高度不等于整棵二叉树的高度-1的话，那么左子树不一定是一棵满二叉树，但是有一点可以确定，
# 右子树一定是一棵满二叉树，这个时候我们就很容易的计算出总结点数nodes=2**(h-1)-1 + 1 +左子树节点数（这里的+1表示root节点）。
# 根据这个思路我们只要不断循环下去直到root==None结束。
class Solution2(object):
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def height(t):
h = -1
while t:
h += 1
t = t.left
return t
h = height(root)
nodes = 0
while root:
if height(root.right) == h-1:
nodes += 2**h
root = root.right
else :
nodes += 2**(h-1)
root = root.left
h -= 1
return nodes
class Solution3(object):
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def getHeight(node):
height = 0
tempNode = node
while tempNode:
height += 1
tempNode = tempNode.left
return height
nodes = 0
tempRoot = root
while tempRoot:
height = getHeight(tempRoot)
rightHeight = getHeight(tempRoot.right)
if rightHeight == height-1:
nodes += 2 ** (height-1)
tempRoot = tempRoot.right
else:
nodes += 2 ** (height-2)
tempRoot = tempRoot.left
return nodes
