Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
Example:
Given a binary tree
1
/ \
2 3
/ \
4 5
Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].
Note: The length of path between two nodes is represented by the number of edges between them.
class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if not root:
return 0
elif not root.right and not root.left:
return 0
elif not root.right:
return max(self.diameterOfBinaryTree(root.left),
1 + self.height(root.left))
elif not root.left:
return max(self.diameterOfBinaryTree(root.right),
1 + self.height(root.right))
else:
return max(self.diameterOfBinaryTree(root.right),
self.diameterOfBinaryTree(root.left),
self.height(root.left) + self.height(root.right) + 2)
def height(self, root):
if not root:
return 0
if not root.right and not root.left:
return 0
return 1 + max(self.height(root.left), self.height(root.right))
class Solution:
def diameterOfBinaryTree(self, root):
self.ans = 0
def depth(node):
if not node:
return 0
r = depth(node.right)
l = depth(node.left)
self.ans = max(self.ans, r+l)
return 1 + max(r, l)
depth(root)
return self.ans