Path Sum
Total Accepted: 81706 Total Submissions: 269391 Difficulty: Easy
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
/**
* 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:
bool hasPathSum(TreeNode* root,int sum,int curSum){
if(!root) return false;
curSum += root->val;
if(root->left==NULL && root->right==NULL){
cout<<curSum<<endl;
if(curSum == sum) return true;
return false;
}
return hasPathSum(root->left,sum,curSum) || hasPathSum(root->right,sum,curSum);
}
bool hasPathSum(TreeNode* root, int sum) {
return hasPathSum(root,sum,);
}
};
Next challenges: (M) Path Sum II (H) Binary Tree Maximum Path Sum (M) Sum Root to Leaf Numbers
Path Sum II
Total Accepted: 65371 Total Submissions: 240118 Difficulty: Medium
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
/**
* 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 {
private:
void pathSum(TreeNode* root, vector<vector<int>>& res, vector<int>& oneRes,int sum,int curSum) {
if(!root) return;
curSum += root->val;
oneRes.push_back(root->val);
if(!root->left && !root->right){
if(curSum == sum){
res.push_back(oneRes);
}
oneRes.pop_back();
return;
}
pathSum(root->left,res,oneRes,sum,curSum);
pathSum(root->right,res,oneRes,sum,curSum);
oneRes.pop_back();
}
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> res;
vector<int> oneRes;
pathSum(root,res,oneRes,sum,);
return res;
}
};
