【easy】437. Path Sum III 二叉树任意起始区间和

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        if (root == NULL)
            return 0;
        return Sum(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum);
    }
private:
    //pre为前面节点的和，cur为前面加上现在遍历到的节点；
    int Sum(TreeNode* root, int pre, int sum){
        if (root == NULL)
            return 0;
        int cur = pre + root->val;

        return (cur == sum) + Sum(root->left, cur, sum) + Sum(root->right, cur, sum);
    }
};

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

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

Return 3. The paths that sum to 8 are:

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

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

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

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

原文地址：https://www.cnblogs.com/sherry-yang/p/8452128.html