题解 | #二叉树中和为某一值的路径(三)#

/**
 * struct TreeNode {
 *  int val;
 *  struct TreeNode *left;
 *  struct TreeNode *right;
 *  TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @param sum int整型
     * @return int整型
     */
    int FindPath(TreeNode* root, int sum) {
        // write code here
        if (root == nullptr) {
            return 0;
        }
        int left = FindPath(root->left, sum);
        int right = FindPath(root->right, sum);
        return left + right + inOrder(root, 0, sum);
    }

    int inOrder(TreeNode* root, int num, int sum) {
        if (root == nullptr) {
            return 0;
        }
        num += root->val;
        int count = (num == sum) ? 1 : 0;
        count += inOrder(root->left, num, sum);
        count += inOrder(root->right, num, sum);
        return count;
    }
};