使用Python查找二叉树中最长连续路径的长度
假设我们有一棵二叉树,我们需要找到其中最长的路径。
因此,如果输入如下:
那么输出将是5,因为最长连续序列是[2, 3, 4, 5, 6]。
为了解决这个问题,我们将采取以下步骤 −
- 如果根为空,则
- 返回0
- maxPath := 0
- 定义一个名为helper()的函数。这将接受节点。
- inc := 1,dec := 1
- 如果节点的左侧不为空,则
- [left_inc,left_dec]:= helper(节点的左侧)
- 否则,
- [left_inc,left_dec]:= [0,0]
- 如果节点的右侧不为空,则
- [right_inc,right_dec]:= helper(节点的右侧)
- 否则,
- [right_inc,right_dec]:= [0,0]
- 如果节点的左侧不为空,且节点的值减去左侧节点的值与1相同,则
- inc:=最大值,使inc和(left_inc +1)
- 否则,当节点的左侧不为空且节点的值减去左侧节点的值与-1相同时,则
- dec:=最大值,使dec和(left_dec +1)
- 如果节点的右侧不为空,并且节点的值减去右侧节点的值与1相同,则
- inc:=最大值,使inc和(right_inc +1)
- 否则,当节点的左侧节点不为空且右侧节点不为空,且左侧节点的值减去节点的值与1相同,右侧节点的值减去节点的值也与1相同时
- maxPath:=最大值,使maxPath和(left_dec + right_inc +1)
- 否则,当节点的左侧节点不为空且右侧节点不为空,且左侧节点的值减去节点的值与-1相同时
- maxPath:=最大值,使maxPath和(left_inc + right_dec +1)
- maxPath:=最大值,使maxPath、inc和dec
- 返回inc、dec
- 从主方法执行以下操作:
- helper(root)
- 返回maxPath
让我们看一下以下实现,以获得更好的理解 −
更多Python相关文章,请阅读:Python 教程
实例
class TreeNode:
def __init__(self, data, left = None, right = None):
self.val = data
self.left = left
self.right = right
def print_tree(root):
if root is not None:
print_tree(root.left)
print(root.val, end = ', ')
print_tree(root.right)
class Solution:
def solve(self, root):
if not root:
return 0
self.maxPath = 0
def helper(node):
inc, dec = 1, 1
if node.left:
left_inc, left_dec = helper(node.left)
else:
left_inc, left_dec = 0, 0
if node.right:
right_inc, right_dec = helper(node.right)
else:
right_inc, right_dec = 0, 0
if node.left and node.val - node.left.val == 1:
inc = max(inc, left_inc + 1)
elif node.left and node.val - node.left.val == -1:
dec = max(dec, left_dec + 1)
if node.right and node.val - node.right.val == 1:
inc = max(inc, right_inc + 1)
elif node.right and node.val - node.right.val == -1:
dec = max(dec, right_dec + 1)
if (node.left and node.right and node.left.val - node.val == 1 and node.val - node.right.val == 1):
self.maxPath = max(self.maxPath, left_dec + right_inc + 1)
elif (node.left and node.right and node.left.val - node.val == -1
and node.val - node.right.val == -1):
self.maxPath = max(self.maxPath, left_inc + right_dec + 1)
self.maxPath = max(self.maxPath, inc, dec)
return inc, dec
helper(root)
return self.maxPath
ob = Solution()
root = TreeNode(3)
root.left = TreeNode(2)
root.right = TreeNode(4)
root.right.left = TreeNode(5)
root.right.right = TreeNode(9)
root.right.left.left = TreeNode(6)
print(ob.solve(root))
输入
root = TreeNode(3)
root.left = TreeNode(2)
root.right = TreeNode(4)
root.right.left = TreeNode(5)
root.right.right = TreeNode(9)
root.right.left.left = TreeNode(6)
输出
5