Path Sum I&&II

 

I

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.

 刚开始想用回溯算法，但是后来发现有负数的情况下这种方法不行，所以就不能用回溯算法了，直接用简单粗暴的递归算法。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool judge(TreeNode *root, int sum,int flag)
    {
       if(root==NULL)
            return false;
       if(root->left==NULL&&root->right==NULL)
            return sum==root->val+flag;
        return judge(root->left,sum,flag+root->val)||judge(root->right,sum,flag+root->val);
    }
    bool hasPathSum(TreeNode *root, int sum) {
        return judge(root,sum,0);
    }
};

　　

Path Sum II

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:
    vector<vector<int>> res;
    vector<int> tempres;
public:
    void subSum(TreeNode* root,int tempSum,int Sum)
    {
        if(root==NULL)
            return ;
        else if((tempSum+root->val==Sum)&&(root->left==NULL&&root->right==NULL))
        {
            tempres.push_back(root->val);
            res.push_back(tempres);
        }
        else
        {
            tempres.push_back(root->val);
            subSum(root->left,tempSum+root->val,Sum);
            subSum(root->right,tempSum+root->val,Sum);
        }
        tempres.pop_back();
        return;
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        if(root==NULL)
            return res;
        else
        {
            subSum(root,0,sum);
            return res;
        }
    }
};