437. Path Sum III
[抄题]:
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
[暴力解法]:
时间分析:
空间分析:
[奇葩输出条件]:
[奇葩corner case]:
[思维问题]:
- 不知道从中间开始截断了应该怎么求。可以分为两种情况:基本的dfs(root求sum)和pathsum(root.left求sum),用两个traverse实现遍历种类的分离,第一次见。
- 对现成的树的研究都是把两个点当参数,做DFS。没有总结思路
[一句话思路]:
count是在符合的基础上计算的,因此累加即可
[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]:
[一刷]:
- dfs的退出条件:不能count = 0时没有执行过就退出了,应该用自增++,扩展时不能执行再自动退出。第一次见。
[二刷]:
[三刷]:
[四刷]:
[五刷]:
[五分钟肉眼debug的结果]:
[总结]:
pathsum传递参数,传递的是sum - root.val,因为之前的已经加过
[复杂度]:Time complexity: O(n) Space complexity: O(n)
[英文数据结构或算法,为什么不用别的数据结构或算法]:
[关键模板化代码]:
if (root.val == sum) { count++; } //execute count += dfs(root.left, sum - root.val); count += dfs(root.right, sum - root.val);
[其他解法]:
[Follow Up]:
[LC给出的题目变变变]:
path sum系列
[代码风格] :
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int pathSum(TreeNode root, int sum) { //corner case if (root == null) { return 0; } return dfs(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum); } public int dfs(TreeNode root, int sum) { //corner case int count = 0; if (root == null) { return 0; } //dfs //exist if (root.val == sum) { count++; } //execute count += dfs(root.left, sum - root.val); count += dfs(root.right, sum - root.val); return count; } }