[LeetCode]437. Path Sum III

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本文介绍了一种解决二叉树中寻找特定和路径的方法。该路径不需要从根节点开始或在叶节点结束，但必须从父节点向下到子节点。通过递归前序遍历算法实现，并使用一个动态数组来维护所有可能的路径和。

题目：

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:

解答：

/**
 * 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) {//每个节点维护一个数组，里面的值是以该节点作为最后一个node，可能的和
        vector<int> maySum;
        return preOrder(root,maySum,sum);

    }
    int preOrder(TreeNode* T,vector<int> maySum,int sum){
        if(T == NULL)
            return 0;
        int result;
        int value = T->val;
        if(value == sum)
            ++result;
        int len = maySum.size();
        for(int i=0;i<len;++i){
            maySum[i] += value;
            if(maySum[i] == sum)
                ++result;
        }
        maySum.push_back(value);

        int leftResult = preOrder(T->left, maySum,sum);
        int rightResult = preOrder(T->right,maySum,sum);
        result += leftResult + rightResult;
        return  result;
        // return result + preOrder(T->left, maySum,sum) + preOrder(T->right,maySum,sum);//这样写超时

    }
};