437. Path Sum III

437. 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

思路：dfs函数，从当前节点依次往下遍历，如果达到目标值，cnt++。dfs2函数：每个节点都要求得该节点的cnt数，最后所有节点的cnt的和就是最后的结果。

/**
* 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 {
void dfs(TreeNode* root, int sum, int &cnt)
{
if(!root) return;
if(root->val==sum) cnt++;
TreeNode* l=root->left,*r=root->right;
dfs(l,sum-root->val,cnt);
dfs(r,sum-root->val,cnt);
}
void dfs2(TreeNode* root,int &sum,vector<int> & ans,int&cnt)
{
if(!root) return;
cnt=0;
dfs(root,sum,cnt);
ans.push_back(cnt);
TreeNode* l=root->left,*r=root->right;
dfs2(l,sum,ans,cnt);
dfs2(r,sum,ans,cnt);
}
public:
int pathSum(TreeNode* root, int sum) {
vector<int> ans;int s=0,cnt=0;
dfs2(root,sum,ans,cnt);
for(int i:ans)
s+=i;
return s;
}
};