Leetcode 437. Path Sum III (二叉树遍历好题)

437. Path Sum III
     Medium
     Given the root of a binary tree and an integer targetSum, return the number of paths where the sum of the values along the path equals targetSum.

The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).

Example 1:

Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
Output: 3
Explanation: The paths that sum to 8 are shown.
Example 2:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: 3

Constraints:

The number of nodes in the tree is in the range [0, 1000].
-109 <= Node.val <= 109
-1000 <= targetSum <= 1000

解法1： 遍历。时间复杂度O(n*n)。

/**
 * 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 targetSum) {
        if (!root) return 0;
        int res = 0;
        helper(root, targetSum, 0, res);
        res += pathSum(root->left, targetSum) + pathSum(root->right, targetSum);
        return res;
    }
private:
    void helper(TreeNode *root, int targetSum, long long sum, int &res) {
        if (!root) return;
        sum += root->val;
        if ((long long)targetSum == sum) {
            res++;
        }
        helper(root->left, targetSum, sum, res);
        helper(root->right, targetSum, sum, res);
    }
};