Leetcode113、路径之和2

1、问题

给定一个二叉树与整数sum,找出所有从根节点到叶结点的路径,这些路 径上的节点值累加和为sum。

2、算法思路

  • 从根节点深度遍历二叉树,先序遍历时,将该节点值存储至path栈中(vector实 现),使用 path_value累加节点值。
  • 当遍历至叶结点时,检查path_value值是否为sum,若为sum,则将path push 进入result结果中。
  • 在后续遍历时,将该节点值从path栈中弹出,path_value减去节点值。

3、代码实现

#include 
#include 

using namespace std;

struct TreeNode
{
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution
{
public:
    vector< vector > pathSum(TreeNode* root, int sum)
    {
        // 求以root为根结点的二叉树中,
        // 哪些路径(从根结点到叶节点)的总和为sum.

        vector< vector > result;  // 保存最终结果
        vector path;  // 单条路径
        int path_value = 0;  // 路径和
        preorder(root, sum, result, path, path_value);

        return result;
    }
private:
    void preorder(TreeNode* node, int sum, vector< vector >& result,
            vector& path, int& path_value)
    {
        if ( !node )
            return;

        // 针对于当前节点,先加
        path_value += node->val;
        // 并保存到路径path中
        path.push_back(node->val);

        // 判断一下
        if (node -> left == NULL && node->right == NULL && path_value == sum)
        {
            result.push_back(path);
        }

        preorder(node->left, sum, result, path, path_value);
        preorder(node->right, sum, result, path, path_value);

        // 左右节点都处理完后
        path_value -= node->val;
        path.pop_back();
    }
};
int main()
{
    TreeNode a(5);
    TreeNode b(4);
    TreeNode c(8);
    TreeNode d(11);
    TreeNode e(13);
    TreeNode f(4);
    TreeNode g(7);
    TreeNode h(2);
    TreeNode i(5);
    TreeNode j(1);

    a.left = &b;
    a.right = &c;
    b.left = &d;
    d.left = &g;
    d.right = &h;
    c.left = &e;
    c.right = &f;
    f.left = &i;
    f.right = &j;

    Solution S;
    vector< vector > result = S.pathSum(&a, 22);
    cout << "total path number: " << result.size() << endl;

    // 包装一下打印结果
    cout << "[" << endl;
    for (int i = 0; i < result.size(); i++)
    {
        cout << "\t[";
        for (int j = 0; j < result[i].size(); j++)
        {
            if (j == (result[i].size() - 1))
                cout << result[i][j];
            else
                cout << result[i][j] << ",";
        }
        if (i == (result.size() - 1))
            cout << "]" << endl;
        else
            cout << "]," << endl;
    }
    cout << "]" << endl;
    return 0;
}

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