Path Sum,Path Sum II

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,0);
    }
};

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,0);
        return res;
    }
};