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, , 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, , sum);
        return ans;
    }
};