Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and doe…

［抄题］: 给出一棵二叉树,寻找一条路径使其路径和最大,路径可以在任一节点中开始和结束(路径和为两个节点之间所在路径上的节点权值之和) ［思维问题］: 不会写分合法 ［一句话思路］: 用两次分治:root2any any2any分一次,左右再分一次. ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: 先root-any左右各一次,再用any-any. ［一刷］: left right都是resultType类型,要用到helper函数…

Given a binary tree, find the maximum path sum. For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root. For exampl…

题目: Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6. 节点可能为负数,寻找一条最路径使得所经过节点和最大.路径可以开始和结束于任何节点但是不能走回头路. 这道题虽然看…

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…

Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 递归求解. maxPathSum(root)跟maxPathSum(root.left)和maxPathSum(roo…

Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6.   用递归确定每一个节点作为root时,从root出发的最长的路径 在每一次递归中计算maxPath   /** * D…

Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example:Given the below binary tree, 1 / \ 2 3 Return 6. 树结构显然用递归来解,解题关键: 1.对于每一层递归,只有包含此层树根节点的值才可以返回到上层.否则路径将不连续. 2.…

124. Binary Tree Maximum Path Sum 题意:给定一个二叉树,每个节点有一个权值,寻找任意一个路径,使得权值和最大,只需返回权值和. 思路:对于每一个节点 首先考虑以这个节点为结尾(包含它或者不包含)的最大值,有两种情况,分别来自左儿子和右儿子设为Vnow. 然后考虑经过这个节点的情况来更新最终答案.更新答案后返回Vnow供父节点继续更新. 代码很简单. 有一个类似的很有趣的题目,给定一个二叉树,选择一条路径,使得权值最大的和最小的相差最大.参考POJ3728 cla…

Binary Tree Maximum Path Sum Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree. For example: Given the below binary tree, 1 / \ 2 3 Return 6. 思想: 后序遍历.注意路径的连通: 结点不为空时要返回  max( max(leftV, rightV)+rootV,…