All parents of a node in BST?

问题

While printing Binary Search Tree(BST) using recursive function (pre-order). I need to print all the parents(path form root) of current node.
An auxiliary data structure can(e.g. path in my code) be use but I don't want to keep node->path to store path.

      4                           
     / \  
    /   \  
   2     6
  / \   / \
 1   3  5  7  

Suppose I am printing nodes in rows using pre-order traverse:

NODE    PATH  
4       4  
2       4,2  
1       4,2,1
3       4,2,3       
6       4,6
5       4,6,5
7       4,6,7  

I did as follows: Working fine!
Path end with 0 (Zero) value in this code. And there is no node value is 0 in BST.

void printpath(int* mypath){
   while(*mypath)  
      printf("%d ", *mypath++);  
}

void preorder(struct tree *p, int* path){
    int *mypath = calloc(sizeof(path)/sizeof(int) + 1 , sizeof(int*));
    int* myp=mypath;

    if(p!=NULL){  
       while( *myp++ = *path++ );  
       --myp;
       *myp=p->data;
       *(myp+1)=0;

        printf("%d PATH ",p->data);
        printpath(mypath);
        printf("\n");
        preorder(p->left, mypath);
        preorder(p->right, mypath);
    }
    free(mypath);
}

But I don't want to keep path array as there is lots of nodes in BST. Can some one suggest me other data-structure/ or method ? A suggestion would be enough but should be efficient.

回答1:

Here's an old trick, which still works: keep the back pointers in the call stack.

    struct stacked_list{
      struct stacked_list* prev;
      struct tree* tree; 
    };

   void printpath_helper(int data, struct stacked_list* path) {
      if (!path->prev)
        printf("%d PATH ", data);
      else
        printpath_helper(data, path->prev);
      printf("%d ", path->tree->data);
    }

    void printpath(struct stacked_list* path) {
      printpath_helper(path->tree->data, path);
      putchar('\n');
    }

    void preorder_helper(struct stacked_list* path) {
      if (path->tree) {
        printpath(path);
        struct stacked_list child = {path, path->tree->left};
        preorder_helper(&child);
        child.tree = path->tree->right;
        preorder_helper(&child);
      }
    }

    void preorder(struct tree* tree) {
      struct stacked_list root = {NULL, tree};
      preorder_helper(&root);
    }

Each recursion of preorder_helper creates an argument struct and passes its address to the next recursion, effectively creating a linked list of arguments which printpath_helper can walk up to actually print the path. Since you want to print the path from top to bottom, printpath_helper needs to also reverse the linked list, so you end up doubling the recursion depth of the function; if you could get away with printing bottom to top, printpath_helper could be a simple loop (or tail recursion).

回答2:

You can use a single array to keep the parents of the current node and pass it down when doing the recursion instead of creating a new array, just remember to resume the array when one recursion finished. I think in that way you can save a lot of memories. The code is below:

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

struct node
{
    int data;
    struct node *left, *right;
};

// A utility function to create a node
struct node* newNode( int data )
{
    struct node* temp = (struct node *) malloc( sizeof(struct node) );

    temp->data = data;
    temp->left = temp->right = NULL;

    return temp;
}



void print_preorder_path(struct node *root,vector<int>& parents)
{
     if(root!=NULL)
     {
              cout<<root->data<<"\t";
              for(size_t i=0;i<parents.size();++i)cout<<parents[i]<<",";
              cout<<root->data<<endl;

              parents.push_back(root->data);
              print_preorder_path(root->left,parents);
              print_preorder_path(root->right,parents);
              parents.pop_back();

     }
}

int main()
{
     // Let us construct the tree given in the above diagram
    struct node *root         = newNode(4);
    root->left                = newNode(2);
    root->left->left          = newNode(1);
    root->left->right         = newNode(3);
    root->right               = newNode(6);
    root->right->left         = newNode(5);
    root->right->right        = newNode(7);

    vector<int> parents;

    cout<<"NODE\tPATH"<<endl;
    print_preorder_path(root,parents);

    getchar();
    return 0;
}

The code is written with stl for simplicity, you can modify as you need, hope it helps!

来源：https://stackoverflow.com/questions/12937475/all-parents-of-a-node-in-bst