把二元查找树转变成排序的双向链表

.把二元查找树转变成排序的双向链表 题目：输入一棵二元查找树，将该二元查找树转换成一个排序的双向链表。要求不能创建任何新的结点，只调整指针的指向。   10 / \ 6 14 / \ / \4 8 12 16 转换成双向链表4=6=8=10=12=14=16。

   
 

void insertOrderToQueue(TN* head,std::queue<TN*>& queue){
    if (!head) {
        return;
    }
    
    if (head -> l) {
        insertOrderToQueue(head -> l,queue);
    }
    
    queue.push(head);
    
    if (head -> r) {
        insertOrderToQueue(head -> r,queue);
    }
  
    
}

TN* search_binary_tree_convert_double_link(TN* head){
    
    if (!head) {
        return NULL;
    }

    std::queue<TN*> queue;
    insertOrderToQueue(head, queue);
    
    if (queue.empty()) {
        return NULL;
    }
    
    TN* doubleListHead = queue.front();
    queue.pop();
    doubleListHead -> l = NULL;
    doubleListHead -> r = NULL;
    
    TN* p =  doubleListHead;
    while (!queue.empty()) {
        
        TN* node = queue.front();
        queue.pop();
        
        p -> r = node;
        node -> l = p;
        
        p = node;
        
    }
    
    p -> r = NULL;
    return doubleListHead;
    /**
     TN* root = NULL;
     int vec[] = {10,8,12,7,9,11,13};
     
     create_TN_tree(root, vec, sizeof(vec) / sizeof(int), 0);
     TN* head = search_binary_tree_convert_double_link(root);
     while (head) {
     NSLog(@"%d ",head -> val);
     head = head -> r;
     }
     */
    
}