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发表于 2023-4-26 19:21:08
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我们需要设计两个算法。第一个算法是按照先序遍历输出森林中每个节点的值及其层次数,第二个算法是求解森林的高度。在此之前,首先我们需要定义森林的数据结构。假设我们的森林由多个二叉树组成,二叉树的结点结构如下:
class TreeNode:
def __init__(self, value):
self.value = value
self.left = None
self.right = None
算法1:按先序次序输出森林中每个结点的值及其对应的层次数
def preorder_traversal(root, level):
if root is None:
return
print("值:", root.value, "层次:", level)
preorder_traversal(root.left, level + 1)
preorder_traversal(root.right, level + 1)
def traverse_forest(forest):
for tree in forest:
preorder_traversal(tree, 1)
算法2:求解森林的高度
def tree_height(root):
if root is None:
return 0
left_height = tree_height(root.left)
right_height = tree_height(root.right)
return max(left_height, right_height) + 1
def forest_height(forest):
max_height = 0
for tree in forest:
tree_h = tree_height(tree)
if tree_h > max_height:
max_height = tree_h
return max_height
示例:
# 创建森林
tree1 = TreeNode(1)
tree1.left = TreeNode(2)
tree1.right = TreeNode(3)
tree1.left.left = TreeNode(4)
tree1.left.right = TreeNode(5)
tree2 = TreeNode(6)
tree2.left = TreeNode(7)
tree2.right = TreeNode(8)
forest = [tree1, tree2]
# 按先序次序输出森林中每个结点的值及其对应的层次数
traverse_forest(forest)
# 求解森林的高度
print("森林的高度:", forest_height(forest))
以上算法可以满足问题的要求。第一个算法实现了按先序次序输出森林中每个结点的值及其对应的层次数,第二个算法计算了森林的高度。 |
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