leetco Path Sum II

和上一题类似，这里是要记录每条路径并返回结果。

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]

]

我们用一个子函数来递归记录，知道叶子节点才判断是否有符合值，有的话就记录。需要注意的是递归右子树之前要把左子树的相应操作去除（见注释）。

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    

    void pathSum(TreeNode *root, vector<vector<int> > &ans, vector<int> tmp, int subsum, int sum)

    {

        if (!root) return ;

        if (!root -> left && !root -> right && subsum + root -> val == sum) 

        {

            tmp.push_back(root -> val);

            ans.push_back(tmp);

        }

        

        if (root -> left)

        {

            tmp.push_back(root -> val);

            subsum += root -> val;

            pathSum(root -> left, ans, tmp, subsum, sum);

            tmp.pop_back(); //因为判断右子树的时候不需要左子树的和

            subsum -= root -> val;

        }

        if (root -> right)

        {

            tmp.push_back(root -> val);

            subsum += root -> val;

            pathSum(root -> right, ans, tmp, subsum, sum);

        }

    }

    vector<vector<int> > pathSum(TreeNode *root, int sum) 

    {

        vector<vector<int> > ans;

        vector<int> tmp;

        

        pathSum(root, ans, tmp, 0, sum);

        return ans;

    }

};

 

其实效率好一些的是对tmp传入引用，例如vector<int> &tmp，那么此时每次记录结果或者左右递归之后都要有一个pop值，来保证tmp符合当前的要求：详见

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

class Solution {

public:

    

    void pathSum(TreeNode *root, vector<vector<int> > &ans, vector<int> &tmp, int subsum, int sum)

    {

        if (!root) return ;

        if (!root -> left && !root -> right && subsum + root -> val == sum) 

        {

            tmp.push_back(root -> val);

            ans.push_back(tmp);

            tmp.pop_back(); // 保持tmp

        }

        

        if (root -> left)

        {

            tmp.push_back(root -> val);

            subsum += root -> val;

            pathSum(root -> left, ans, tmp, subsum, sum);

            tmp.pop_back(); // 因为判断右子树的时候不需要左子树的和

            subsum -= root -> val; // 同上理

        }

        if (root -> right)

        {

            tmp.push_back(root -> val);

            subsum += root -> val;

            pathSum(root -> right, ans, tmp, subsum, sum);

            tmp.pop_back(); // 保持tmp

        }

    }

    vector<vector<int> > pathSum(TreeNode *root, int sum) 

    {

        vector<vector<int> > ans;

        vector<int> tmp;

        

        pathSum(root, ans, tmp, 0, sum);

        return ans;

    }

};