从通用树或N元树中删除所有叶节点

在通用树或N元树中删除所有叶节点是一个常见的操作。下面给出C++实现：

#include <iostream>
#include <vector>

using namespace std;

class TreeNode {
public:
    int val;
    vector<TreeNode*> children;
    TreeNode(int val) : val(val) {};
};

TreeNode* deleteLeaves(TreeNode* root) {
    if (!root) return NULL;
    if (root->children.empty()) {
        delete root;
        return NULL;
    }
    for (auto& child : root->children) {
        child = deleteLeaves(child);
    }
    root->children.erase(
        remove_if(
            root->children.begin(),
            root->children.end(),
            [](TreeNode* child){ return !child; }
        ),
        root->children.end()
    );
    if (root->children.empty()) {
        delete root;
        return NULL;
    }
    return root;
}

int main() {
    TreeNode* root = new TreeNode(1);
    TreeNode* child1 = new TreeNode(2);
    TreeNode* grandchild1 = new TreeNode(3);
    TreeNode* grandchild2 = new TreeNode(4);
    TreeNode* child2 = new TreeNode(5);
    TreeNode* grandchild3 = new TreeNode(6);
    TreeNode* grandchild4 = new TreeNode(7);
    root->children.push_back(child1);
    child1->children.push_back(grandchild1);
    child1->children.push_back(grandchild2);
    root->children.push_back(child2);
    child2->children.push_back(grandchild3);
    child2->children.push_back(grandchild4);
    root = deleteLeaves(root);
    delete root;
    return 0;
}

这里给出函数deleteLeaves的实现。如果节点为空，则直接返回空指针；如果节点没有子节点，则删除这个节点并返回空指针。否则，依次递归删除子节点，并从子节点列表中删除空指针。最后，如果节点没有子节点，则删除这个节点并返回空指针。最终，函数返回根节点。

这段代码的时间复杂度为$O(N)$，其中$N$为节点数。空间复杂度为$O(H)$，其中$H$为树的高度。