题解 | #二叉树中和为某一值的路径(三)#

class Solution {
public:
     int FindPath(TreeNode* root, int sum, bool allowBreak) {
        if (!root) return 0;
        int count = sum == root->val ? 1 : 0;
        if (!root->left && !root->right) {
            return count;
        }

        auto left = root->left;
        if (left) {
            if (allowBreak) {
                count += FindPath(left, sum, true); 
            }
            count += FindPath(left, sum - root->val, false);
        }

        auto right = root->right;
        if (right) {
            if (allowBreak) {
                count += FindPath(right, sum, true);
            }
            count += FindPath(right, sum - root->val, false);
        }
        return count;
    }
    int FindPath(TreeNode* root, int sum) {
        return FindPath(root, sum, true);
    }
};