标签：div which more com style 时间 code path sum example
https://leetcode.com/problems/path-sum-iii
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
解题思路：
这题虽然简单，写对了可不容易，想写出线性的方法更不容易。
简单的解法，一个个节点看。每个节点都往下找他的所有子节点，路径和为sum的，就算一个。
这样做的时间复杂度是O(n^2)。
/** * 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) { int[] count = new int[1]; count(root, sum, count); if (root != null) { count[0] += pathSum(root.left, sum); count[0] += pathSum(root.right, sum); } return count[0]; } public void count(TreeNode root, int sum, int[] count) { if (root == null) { return; } if (root.val == sum) { count[0]++; } if (root.left != null) { count(root.left, sum - root.val, count); } if (root.right != null) { count(root.right, sum - root.val, count); } } }
标签：div which more com style 时间 code path sum example
原文地址：https://www.cnblogs.com/NickyYe/p/9972448.html
