求完全二叉树节点个数

``````#(0,0,0),(0,0,1)(0,1,0)(0,1,1)(1,0,0)(1,0,1)(1,1,0)(1,1,1)其实是
0，1，2，3，4，5，6，7
``````
``````# 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):
# 先算数的深度
def findlevel(self, root):
if root:
return 1+self.findlevel(root.left)
else:
return 0
'''step就是用1代表(0,0,1),3'''
def findendNode(self, root, step, level):
#print step,level
temp = root
for i in range(level):
if ((step&(2**(level-1-i))) >> (level-1-i)) == 0:
#print 'xxx'
temp = temp.left
else:
temp = temp.right
if temp:
return True
else:
return False

'''
def findendNode(self, root, step, level):
temp = root
lr = list(bin(step))[2:]
#print 'lr', lr, 'level', level
lrlen = len(lr)
# 左子树
for i in range(level-lrlen):
temp = temp.left
for i in range(lrlen):
if lr[i] == '0':
temp = temp.left
else:
temp = temp.right
if temp:
return True
else:
return False
'''
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return 0
level=self.findlevel(root)-1
if level == 0:
return 1
endend = 2**level
s = endend - 1
# 找出深度后呢，找出结尾，看下结尾排列，看下h = 3最后一排的排列,0代表向左子树搜索，1代表向右子树搜索，可以用bit特         # 位表示，可以用二分查找法
#(0,0,0),(0,0,1)(0,1,0)(0,1,1)(1,0,0)(1,0,1)(1,1,0)(1,1,1)其实就是0，1，2，3，4，5，6，7
start = 0
end = endend-1
while True:
# 从中间找
temp = (end+start)/2
if not self.findendNode(root, temp, level):
# 如果是false，设置为end
end = temp
else:
# 如果是true，设置为起点
start = temp
if end-start == 1:
if end == endend-1:
if self.findendNode(root, end, level):
s += endend
return s
else:
s+= end
return s
else:
s += end
return s
``````