题目描述
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.
例子
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:
- 5 -> 3
- 5 -> 2 -> 1
- -3 -> 11
解题思路
方法一
可以用一个变量来维护从 root
到当前节点的和,再递归的以所有节点为 root
进行遍历,代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int sum) {
if (root == nullptr) return 0;
return pathSumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
}
private:
int pathSumUp(TreeNode* node, int current_sum, int target_sum) {
if (node == nullptr) return 0;
current_sum += node->val;
return (current_sum == target_sum) + pathSumUp(node->left, current_sum, target_sum) + pathSumUp(node->right, current_sum, target_sum);
}
};
- 时间复杂度: O(n^2)
- 空间复杂度: O(n)
方法二
第一种方法本质上可以转换成判断两条路径,一个是从 root
到 节点 i
, 一条是从 root
到节点 j
(必须要经过 i
),计算这两条路径的和(前缀和 prefix
)并相减就可以知道从 i
到 j
的和;根据这个思路,如果我们只用一次遍历树,并记录从 root
到当前节点 i
的和,那么问题就转换成,在 j
之前有多少个节点 i
满足 prefix(i) = prefix(j) - target
,根据这个思路,我们可以用一个哈希表来存储所有节点的 prefix
,这样就不需要递归遍历,注意一下当遍历完当前子树的时候应该将当前子树的所有 prefix
从哈希表删掉,不然会影响其他子树的结果,代码如下:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int pathSum(TreeNode* root, int sum) {
ans = 0;
prefix_sum_count[0] = 1;
dfs(root, 0, sum);
return ans;
}
private:
std::unordered_map<int, int> prefix_sum_count;
int ans;
void dfs(TreeNode* node, int current_sum, int target_sum) {
if (node == nullptr) return;
current_sum += node->val;
ans += prefix_sum_count[current_sum - target_sum];
prefix_sum_count[current_sum]++;
dfs(node->left, current_sum, target_sum);
dfs(node->right, current_sum, target_sum);
// key step!
prefix_sum_count[current_sum]--;
}
};
- 时间复杂度: O(n)
- 空间复杂度: O(h) -> 只跟树的最大高度有关,因为哈希表不会同时存放左右两个子树的值