543 Diameter of Binary Tree - Easy

Problem:

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

Problem Analysis:

• use dfs to get O(n) time complexity
• a little tricky with the add 2

Solutions:

# 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:
        res = [0]

        def dfs(root):
            if not root:
                return -1
            left = dfs(root.left)
            right = dfs(root.right)
            # add 2 cos the root is -1 to balance back
            res[0] = max(res[0], 2+left+right)

            return 1+ max(left, right)
        
        dfs(root)
        return res[0]

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