Given the root of a binary tree, return 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.
The length of a path between two nodes is represented by the number of edges between them.
Example 1:
Input: root = [1,2,3,4,5] Output: 3 Explanation: 3 is the length of the path [4,2,1,3] or [5,2,1,3].
Example 2:
Input: root = [1,2] Output: 1
- The number of nodes in the tree is in the range
[1, 104]. -100 <= Node.val <= 100
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution:
def diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:
max_diameter = [0] # Using a list to allow modification within the nested function
def depth(node, max_diameter):
if not node:
return 0
left_depth = depth(node.left, max_diameter)
right_depth = depth(node.right, max_diameter)
# Update the maximum diameter
max_diameter[0] = max(max_diameter[0], left_depth + right_depth)
return 1 + max(left_depth, right_depth)
depth(root, max_diameter)
return max_diameter[0]The time complexity is O(n), where n is the number of nodes in the tree. This is because we visit each node exactly once in the recursive traversal.
The space complexity is O(h), where h is the height of the tree. This is due to the recursive call stack. In the worst case, the tree could be linear (i.e., a linked list), in which case the height of the tree would be equal to the number of nodes, leading to a space complexity of O(n).