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:Given a binary t…

题目: 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:Given a bina…

给定一棵二叉树,你需要计算它的直径长度.一棵二叉树的直径长度是任意两个结点路径长度中的最大值.这条路径可能穿过根结点.示例 :给定二叉树          1         / \        2   3       / \           4   5    返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3].注意:两结点之间的路径长度是以它们之间边的数目表示.详见:https://leetcode.com/problems/diameter-of-binary-t…

题目中的直径定义为: 任意两个节点的最远距离 没想出来,看的答案 思路是:diameter = max(左子树diameter,右子树diameter,(左子树深度+右子树深度+1)) 遍历并更新结果 int res = 1; public int diameterOfBinaryTree(TreeNode root) { helper(root); return res-1; } public int[] helper(TreeNode root) { //两个量分别是节点深度,节点最大dia…

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:Given a binary t…

［抄题］: 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:Given a bi…

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 longestpath between any two nodes in a tree. This path may or may not pass through the root. Example:Given a binary tr…

给定一棵二叉树,你需要计算它的直径长度.一棵二叉树的直径长度是任意两个结点路径长度中的最大值.这条路径可能穿过根结点. 示例 : 给定二叉树 1 / \ 2    3 / \ 4  5 返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3]. 注意:两结点之间的路径长度是以它们之间边的数目表示. class Solution { public: int res = 0; int diameterOfBinaryTree(TreeNode* root) { if(root ==…

124. Binary Tree Maximum Path Sum https://www.cnblogs.com/grandyang/p/4280120.html 如果你要计算加上当前节点的最大path和,这个节点的左右子树必定是纯左右树(即没有拐点), 用另一个参数保留整个二叉树的最大path和,然后计算每一个以当前节点为拐点的路径和并且不断更新整个二叉树的最大值 函数的返回值是纯左右子树的最大path,没有拐点 这个题目给root定位为非空,所以直接这样写可以.如果root为空,这样写就会…

problem 543. Diameter of Binary Tree 题意: 转换一种角度来看,是不是其实就是根结点1的左右两个子树的深度之和呢.那么我们只要对每一个结点求出其左右子树深度之和,这个值作为一个候选值,然后再对左右子结点分别调用求直径对递归函数,这三个值相互比较,取最大的值更新结果res,因为直径不一定会经过根结点,所以才要对左右子结点再分别算一次.为了减少重复计算,我们用哈希表建立每个结点和其深度之间的映射,这样某个结点的深度之前计算过了,就不用再次计算了. solution…