LeetCode437 Path Sum III

文章目录
  1. 1. 描述
  2. 2. 样例
  3. 3. 思路
  4. 4. 代码

描述

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.

样例

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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

思路

求二叉树中有多少条权值和为sum的路径。

遍历每个结点,以该结点作为根结点DFS搜索答案。

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
// 以root为根结点,路径和为sum的数目
int find(TreeNode* root, int sum) {
if (root == NULL) return 0;
return find(root->left, sum - root->val) // 找左儿子中权值和为(sum - root->val)的路径数
+ find(root->right, sum - root->val)
+ (sum == root->val); // 相等说明找到一条路径
}
int pathSum(TreeNode* root, int sum) {
if (root == NULL) return 0;
return find(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
};
分享到 评论