题解 | #二叉树中和为某一值的路径(二)#

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
  public:
    vector<vector<int>> FindPath(TreeNode* root, int expectNumber) {
		result.clear();
		path.clear();
		if (root == nullptr) return result;
		path.push_back(root->val);
		traversal(root, expectNumber - root->val);
		return result;
    }
    void traversal(TreeNode* cur, int count) {
        // 叶子节点且找到了和为sum的路径
        if (cur->left == nullptr && cur->right == nullptr && count == 0) {
            result.push_back(path);
            return;
        }
        if (cur->left == nullptr && cur->right == nullptr && count != 0) {
            return;
        }
        if (cur->left) {
            path.push_back(cur->left->val);
            count -= cur->left->val;
            traversal(cur->left, count);
            count += cur->left->val;
            path.pop_back();
        }

        if (cur->right) {
            path.push_back(cur->right->val);
            count -= cur->right->val;
            traversal(cur->right, count);
            count += cur->right->val;
            path.pop_back();
        }
		return;
    }
  private:
    vector<int> path;
    vector<vector<int>> result;
};