543-diameter-of-binary-tree

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Description

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

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -100 <= Node.val <= 100

Solution(Python)

# Definition for a binary tree node.
# 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:
        return self.optimize(root)

    # Time Complexity: O(n^2)
    # Space Complexity: O(H)
    def bruteforce(self, node: Optional[TreeNode]) -> int:
        if node is None:
            return 0

        lheight = self.height(node.left)
        rheight = self.height(node.right)

        ldiameter = self.bruteforce(node.left)
        rdiameter = self.bruteforce(node.right)

        return max(lheight + rheight, max(ldiameter, rdiameter))

    # Time Complexity: O(n)
    # Space Complexity: O(H)
    def optimize(self, node: Optional[TreeNode]) -> int:
        ans = 0

        def dfs(node):
            nonlocal ans
            if node is None:
                return 0

            left_height = dfs(node.left)
            right_height = dfs(node.right)

            cur_width = left_height + right_height

            if cur_width > ans:
                ans = cur_width

            return 1 + max(left_height, right_height)

        dfs(node)
        return ans

    def height(self, node):
        if node is None:
            return 0
        return 1 + max(self.height(node.left), self.height(node.right))