Count complete tree nodes¶
Time: O((LogN)^2); Space: O(1); medium
Given a complete binary tree, count the number of nodes.
Notes:
Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.
Example 1:
Input: root = {TreeNode} [1,2,3,4,5,6]
1
/ \
2 3
/ \ /
4 5 6
Output: 6
Example 2:
Input: root = {TreeNode} [1,2,3,4,5,6,8]
1
/ \
2 3
/ \ /\
4 5 6 8
Output: 7
[8]:
class TreeNode(object):
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
1. Binary Search [O((LogN)^2), O(1)]¶
[9]:
class Solution1(object):
"""
Time: O(H*LogN)=O((LogN)^2)
Space: O(1)
"""
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root is None:
return 0
node, level = root, 0
while node.left is not None:
node = node.left
level += 1
# Binary search.
left, right = 2 ** level, 2 ** (level + 1)
while left < right:
mid = left + (right - left) // 2
if not self.exist(root, mid):
right = mid
else:
left = mid + 1
return left - 1
# Check if the Nth node exist
def exist(self, root, n):
k = 1
while k <= n:
k <<= 1
k >>= 2
node = root
while k > 0:
if (n & k) == 0:
node = node.left
else:
node = node.right
k >>= 1
return node is not None
[10]:
s = Solution1()
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left = TreeNode(6)
assert s.countNodes(root) == 6
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left, root.right.right = TreeNode(6), TreeNode(8)
assert s.countNodes(root) == 7
2. Binary Search [O((LogN)^2), O(1)]¶
[11]:
class Solution2(object):
"""
Time: O(H*H) = O((LogN)^2)
Space: O(1)
"""
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def height(root):
h = -1
while root:
h += 1
root = root.left
return h
result, h = 0, height(root)
while root:
if height(root.right) == h-1:
result += 2**h
root = root.right
else:
result += 2**(h-1)
root = root.left
h -= 1
return result
[12]:
s = Solution2()
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left = TreeNode(6)
assert s.countNodes(root) == 6
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left, root.right.right = TreeNode(6), TreeNode(8)
assert s.countNodes(root) == 7
3. Binary Search [O((LogN)^2), O(1)]¶
[13]:
class Solution3(object):
"""
Time: O(H * LogN) = O((LogN)^2)
Space: O(1)
"""
def countNodes(self, root):
"""
:type root: TreeNode
:rtype: int
"""
def check(node, n):
base = 1
while base <= n:
base <<= 1
base >>= 2
while base:
if (n & base) == 0:
node = node.left
else:
node = node.right
base >>= 1
return bool(node)
if not root:
return 0
node, level = root, 0
while node.left:
node = node.left
level += 1
left, right = 2**level, 2**(level+1)-1
while left <= right:
mid = left+(right-left) // 2
if not check(root, mid):
right = mid - 1
else:
left = mid + 1
return right
[14]:
s = Solution3()
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left = TreeNode(6)
assert s.countNodes(root) == 6
root = TreeNode(1)
root.left, root.right = TreeNode(2), TreeNode(3)
root.left.left, root.left.right = TreeNode(4), TreeNode(5)
root.right.left, root.right.right = TreeNode(6), TreeNode(8)
assert s.countNodes(root) == 7