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 ...
也可以在迭代每一个node的时候,记录下其diameter,ans就是所有node diameter最大值: #Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = NoneclassSolution(object):defdiameterOfBinaryTree(self, root):""":type root:...
Given a binary tree, you need to compute 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. Example: Givenabinary tree1/ \23/ \45Return3, which...
题目原址 https://leetcode.com/problems/diameter-of-binary-tree/description/ 题目描述 Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the lengt...LeetCode 543. Diameter of Binary Tree Description Given a binary tree, you...
The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path ...leetcode 543. Diameter of Binary Tree Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of ...
Can you solve this real interview question? Diameter of Binary Tree - 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 p
Leetcode-Easy 543. Diameter of Binary Tree 543. Diameter of Binary Tree 描述: 求二叉树最长路径长度 思路: 深度优先搜索 代码 # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None...
leetcode-543-Diameter of Binary Tree int dfs(root): the max. length of one of substree. i.e. dfs(root): if(root == null) return 0 left += dfs(root->left) right += dfs(root->right) res = max(res, left + right + 1);...
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