利用递归函数的返回值

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;

        if (root == NULL)
            return res;

        if (root->left == NULL && root->right == NULL) {
            res.push_back( to_string(root->val));
            return res;
        }

        vector<string> leftS = binaryTreePaths(root->left);
        for (int i = 0; i < leftS.size(); i++) {
            res.push_back( to_string(root->val) + "->" + leftS[i]);
        }
        
        vector<string> rightS = binaryTreePaths(root->right);
        for (int i = 0; i < rightS.size(); i++) {
            res.push_back( to_string(root->val) + "->" + rightS[i]);
        }
        return res;
    }
};

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

      10
     /  \
    5   -3
   / \    \
  3   2   11
 / \   \
3  -2   1

返回 3。和等于 8 的路径有:

1.  5 -> 3
2.  5 -> 2 -> 1
3.  -3 -> 11

class Solution {
public:
    int ans=0;
    int pathSum(TreeNode* root, int sum) {
        if(root){
            dfs(root,sum);
            pathSum(root->left,sum);
            pathSum(root->right,sum);
        }
        return ans;
    }
    void dfs(TreeNode* root, int sum){
        if(!root)return;
        if(root->val==sum)ans++;
        dfs(root->left,sum-root->val);
        dfs(root->right,sum-root->val);
    }
};

// @lc code=start
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        
        if ( root == NULL )
            return 0;
        int res = findPath( root, sum);
        res += pathSum( root->left, sum);
        res += pathSum( root->right, sum);

        return res;
    }

private:
    // 在以node为根节点的二叉树中，寻找包含node的路径，和为sum
    // 返回这样的路径个数
    int findPath( TreeNode* node, int num) {

        if ( node == NULL)
            return 0;
        int res = 0;
        if( node->val == num)
            res += 1;
        
        res += findPath(node->left, num - node->val);
        res += findPath(node->right, num - node->val);

        return res;
    }
};
// @lc code=end