计算二叉树中距离根节点最远的叶子节点的个数,即二叉树中最后一层的叶子节点的个数。
如在下面的二叉树中,叶子节点的总数量为 4(分为3、4、6、8),最后一层的叶子节点的个数为 1(编号8):
1
/ \
2 3
/ \
4 5
/ \
6 7
\
8
方法:层序遍历
code:
from collections import deque
class Node:
def __init__(self, value, left=None, right=None):
self.value = value
self.left = left
self.right = right
# 计算二叉树中最后一层的叶子节点的个数
def leafNodeNum(root):
if not root:
return 0
num = 0
stack = deque()
stack.append(root)
# 层序遍历
while stack:
# 当前层级的叶子节点个数
cur_num = 0
length = len(stack)
for _ in range(length):
# 必须使用 popleft() 保证先进先出,不能使用 pop()
# 使用 length 保证将上一层的节点全部弹出,而不弹出当前层级新加入的节点
cur_node = stack.popleft()
# 判断当前节点是否为叶子节点
if not cur_node.left and not cur_node.right:
cur_num += 1
if cur_node.left:
stack.append(cur_node.left)
if cur_node.right:
stack.append(cur_node.right)
# 更新最大叶子节点个数
num = max(num, cur_num)
return num
if __name__ == '__main__':
# 构建二叉树
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.left.right.left = Node(6)
root.left.right.right = Node(7)
root.left.right.right.left = Node(8)
# 计算二叉树距离根节点最远的叶子节点个数
max_leaves_count = leafNodeNum(root)
print("二叉树中最后一层的叶子节点的数量为:", max_leaves_count)