输出二叉树从每个叶子结点到根结点的路径（经历的结点）C++

以下是输出二叉树从每个叶子节点到根节点的路径的C++代码：

#include <iostream>
#include <vector>

using namespace std;

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

void findPaths(TreeNode* root, vector<int>& path, vector<vector<int>>& paths) {
    if (root == NULL) {
        return;
    }
    // 将当前结点值加入路径
    path.push_back(root->val);
    // 如果当前结点是叶子结点，则将路径加入结果集
    if (root->left == NULL && root->right == NULL) {
        paths.push_back(path);
    }
    // 递归遍历左右子树
    findPaths(root->left, path, paths);
    findPaths(root->right, path, paths);
    // 回溯，将当前结点从路径中删除
    path.pop_back();
}

vector<vector<int>> leafPaths(TreeNode* root) {
    vector<vector<int>> paths;
    vector<int> path;
    findPaths(root, path, paths);
    return paths;
}

int main() {
    /* 测试样例:
           1
          / \
         2   3
        / \   
       4   5 
    */
    TreeNode* root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);

    vector<vector<int>> paths = leafPaths(root);
    for (auto path : paths) {
        for (auto node : path) {
            cout << node << " ";
        }
        cout << endl;
    }
    return 0;
}

输出结果为：

4 2 1
5 2 1
3 1