二叉树中的最大路径和 · Binary Tree Maximum Path Sum

[抄题]:

[思维问题]:

[一句话思路]:

用两次分治。

[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]:

[二刷]:

[三刷]:

[四刷]:

[五刷]:

[总结]:

[复杂度]:Time complexity: O(n) Space complexity: O()

[英文数据结构,为什么不用别的数据结构]:

[其他解法]:

暴力解法 把所有路径找一遍:n^2

[Follow Up]:

root-leaf 就算有负数也得走:直接左右合并

root-any node 有负数可以不走:max(0,max(left,right)) + root.val 结果要和0比,小于0就只有root.val得了

[LC给出的题目变变变]:

原文地址:https://www.cnblogs.com/immiao0319/p/8376474.html

时间: 2024-10-11 20:00:14

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