437. Path Sum III

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本文介绍了一个在二叉树中寻找路径和等于指定值的有效算法。该算法无需从根节点开始，也不必在叶子节点结束，路径只需沿着父节点指向子节点的方向进行。文中提供了一个具体的示例，展示了如何在一个给定的二叉树中找到所有和为8的路径。

题目描述【Leetcode】

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.

Example:
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:

5 -> 3
5 -> 2 -> 1
-3 -> 11
求满足路径和为sum的路径条数：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
int f(TreeNode* root, int sum) {
        if(!root) return 0;
        int count = 0;
        if(root->val == sum) count++;
        count += f(root->left,sum-root->val)+f(root->right,sum-root->val);
        return count;
    }

class Solution {
public:
    int pathSum(TreeNode* root, int sum) {
        int count = 0;
        if(!root) return 0;
        count = count + f(root,sum)+ pathSum(root->left,sum)+pathSum(root->right,sum);         
        return count;
    }
};