[leetcode] Path sum路径之和

要求
给定树，与路径和，判断是否存在从跟到叶子之和为给定值的路径。比如下图中，给定路径之和为22，存在路径<5,4,11,2>，因此返回true;否则返回false.
              5

             /             4   8

           /   /           11  13  4

         /  \              7    2      5
思路
递归，从跟到叶子判断，如果在叶子处剩下的给定值恰好为给定值，那么返回ture.
参考代码

/**

 * 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 hasPathSum(TreeNode *root, int sum) {

        if (root == NULL)

            return false;

        if (root->left == NULL && root->right == NULL && root->val == sum)

            return true;

        if (root->left != NULL && hasPathSum(root->left, sum - root->val))

            return true;

        if (root->right != NULL && hasPathSum(root->right, sum - root->val))

            return true;

        return false;

    }

};

扩展

求出所有符合条件的路径。例如，上题中返回<<5, 4, 11, 2>, <5, 8, 4, 5>>

参考代码

/**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

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

 * };

 */

class Solution {

public:

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

        vector<vector<int> > res;

        vector<int> tmp;

        if (root == NULL)

            return res;

        tmp.push_back(root->val);

        pathSumRecur(root, sum, tmp, res);

        return res;

    }

    void pathSumRecur(TreeNode *root, int sum , vector<int> &tmp, vector<vector<int> > &res)

    {

        if (root == NULL)

            return;

        if (root->left == NULL && root->right == NULL && root->val == sum)

        {

            res.push_back(tmp);

        }

        if(root->left != NULL)

        {

            tmp.push_back(root->left->val);

            pathSumRecur(root->left, sum - root->val, tmp, res);

            tmp.pop_back();

        }

        if(root->right != NULL)

        {

            tmp.push_back(root->right->val);

            pathSumRecur(root->right, sum - root->val, tmp, res);

            tmp.pop_back();

        }

    }

};