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Copy path1367. Linked List in Binary Tree.py
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1367. Linked List in Binary Tree.py
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class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right) class Solution(object):
# This function checks whether there is a path in the binary tree that matches the current linked list path.
def checkPath(self, head, root):
# If we've reached the end of the linked list, it means the path matches, so return True.
if not head:
return True
# If we've reached a leaf node in the binary tree (root is None), or if the current values of the linked list
# and the binary tree node don't match, return False.
if not root or head.val != root.val:
return False
# Recursively check the left and right child of the binary tree to see if the next node in the linked list
# can be found in either subtree.
return self.checkPath(head.next, root.left) or self.checkPath(head.next, root.right)
def isSubPath(self, head, root):
# If the binary tree is empty, return False because there can't be a subpath.
if not root:
return False
# Check if the current binary tree node matches the linked list's head and, if so, check the path from there.
# Also, recursively check the left and right subtrees to see if the linked list is a subpath starting
# from those nodes.
return (head.val == root.val and self.checkPath(head, root)) or self.isSubPath(head, root.left) or self.isSubPath(head, root.right)