Leetcode 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
/**
* 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:
//不是路径第一个节点 不能断开
void funpath(TreeNode* root, int sum,int&ans){
if(root){
if(root->val==sum)ans++;
funpath(root->left,sum-root->val,ans);
funpath(root->right,sum-root->val,ans);
}
}
//root为路径的第一个节点
void funroot(TreeNode* root, int sum,int&ans){
if(root){
if(root->val==sum)ans++;
funpath(root->left,sum-root->val,ans);//如果前面已经成功 后面和为0就ojbk了
funpath(root->right,sum-root->val,ans);
funroot(root->left,sum,ans);
funroot(root->right,sum,ans);
}
}
int pathSum(TreeNode* root, int sum) {
int ans=0;
funroot(root,sum,ans);
return ans;
}
};
Success
Details
Runtime: 24 ms, faster than 75.74% of C++ online submissions for Path Sum III.
Memory Usage: 15.6 MB, less than 88.99% of C++ online submissions for Path Sum III.