Reputation: 9064
I'm wondering how I can optimally find the diameter (or longest path between any two leaf nodes) of a binary tree. I have the basic solution below, but the second solution requires passing pointers. How can I do something like this in Python?
def find_tree_diameter(node):
if node == None:
return 0
lheight = height(node.left)
rheight = height(node.right)
ldiameter = find_tree_diameter(node.left)
rdiameter = find_tree_diameter(node.right)
return max(lheight+rheight+1, ldiameter, rdiameter)
def find_tree_diameter_optimized(node, height):
lheight, rheight, ldiameter, rdiameter = 0, 0, 0, 0
if node == None:
# *height = 0;
return 0
ldiameter = diameterOpt(root.left, &lheight)
rdiameter = diameterOpt(root.right, &rheight)
# *height = max(lheight, rheight) + 1;
return max(lh + rh + 1, max(ldiameter, rdiameter));
Upvotes: 2
Views: 2273
Reputation: 328
Simple Python 3 solution
def findDepth(root):
if root is None:
return 0
return 1 + max(findDepth(root.left), findDepth(root.right))
class Solution:
def diameterOfBinaryTree(self, root: TreeNode) -> int:
if root is None:
return 0
left = findDepth(root.left)
right = findDepth(root.right)
ldia = self.diameterOfBinaryTree(root.left)
rdia = self.diameterOfBinaryTree(root.right)
return max(left+right, max(ldia, rdia))
Upvotes: 1
Reputation: 58319
Python supports multiple return values, so you don't need pointer arguments like in C or C++. Here's a translation of the code:
def diameter_height(node):
if node is None:
return 0, 0
ld, lh = diameter_height(node.left)
rd, rh = diameter_height(node.right)
return max(lh + rh + 1, ld, rd), 1 + max(lh, rh)
def find_tree_diameter(node):
d, _ = diameter_height(node)
return d
The function diameter_height
returns the diameter and the height of the tree, and find_tree_diameter
uses it to just compute the diameter (by discarding the height).
The function is O(n), no matter the shape of the tree. The original function is O(n^2) in the worst case when the tree is very unbalanced because of the repeated height calculations.
Upvotes: 12