[LeetCode]437. Path Sum III

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本文介绍了一道经典的二叉树题目——路径总和III的解法。该问题要求在给定的二叉树中找到所有节点路径之和等于特定值的路径数量，路径可以始于任意节点，并向下遍历到任意子节点。通过递归算法实现，首先定义了一个辅助函数来计算从当前节点开始的路径总和是否符合要求，再递归地计算左子树和右子树中符合条件的路径数量。

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437. Path Sum III

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

/**
 * 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) {
        int result = 0;
        if(root == NULL)
            return result;
        find(root, sum, result);
        return result + pathSum(root->left, sum)+pathSum(root->right, sum);
    }

    // 给我一个起点，找出以这个点为起点的所有可能路径数
    void find(TreeNode* root, int target, int& result){
        if(root == NULL)
            return;
        if(root->val == target)
            ++result;
        if(root->left)
            find(root->left, target-root->val, result);
        if(root->right)
            find(root->right, target-root->val, result);
    }
};