C/C++教程

[Leetcode] 543. Diameter of Binary Tree_Easy Tag: DFS

本文主要是介绍[Leetcode] 543. Diameter of Binary Tree_Easy Tag: DFS,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
543. Diameter of Binary Tree Easy

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

 

Ideas: 最长一条边的node的个数 - 1, 建立一个helper function, get the max number of nodes with the root, 再用self.an = max(self.ans, left + right + 1 - 1) 去得到边

Code

# 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: TreeNode) -> int:
        self.ans = 0
        self.helperWithRoot(root)
        return self.ans
    
    # get the number of nodes 
    def helperWithRoot(self, root):
        if not root: return 0
        left = self.helperWithRoot(root.left)
        right = self.helperWithRoot(root.right)
        self.ans = max(self.ans, left + right )
        return 1 + max(left, right)

 

这篇关于[Leetcode] 543. Diameter of Binary Tree_Easy Tag: DFS的文章就介绍到这儿,希望我们推荐的文章对大家有所帮助,也希望大家多多支持为之网!