二叉树的直径

1、题目

图1

2、题解

深度优先搜索

递归计算当前节点的左右子树的路径，当前节点的直径值为\(max(左子树路径,右子树路径)+1\)。递归至树节点，该树的直径为左子树深度+右子树深度+1。

# 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:
        self.ans = 1

        def dfs(node):
            if not node: return 0

            L = dfs(node.left)
            R = dfs(node.right)
            self.ans = max(self.ans, L+R+1)

            return max(L,R)+1

        dfs(root)
        
        return self.ans - 1