在伪二叉树中为每个子节点设置兄弟节点

Question

could anyone tell me what is the simplest algorithm with use of recursion which would take a root of a so called binary tree (so called because it is not strictly speaking binary tree) and make every child in this tree connected with its sibling.
So if I have:

             1
           /   \  
          2     3  
         / \     \  
        4   5     6
       /           \
      7             8  

then the sibling to 2 would be 3, to four five, to five six and to seven eight.

Answer 1

Do a BFS, assign level numbers to nodes and connect nodes with the same level number.

Pseudocode:

void connectSiblings(Node root)
{
    Queue q = new Queue();
    root.level = 1;
    q.enqueue(root)
    while(!q.isEmpty())
    {
        elem = q.dequeue();
        //add elem's children to q
        if (elem.left != NULL)
        {
            elem.left.level = elem.level + 1;
            q.enqueue(elem.left);
        }
        if (elem.right != NULL)
        {
            elem.right.level = elem.level + 1;
            q.enqueue(elem.right);
        }

        //check level numbers and assign siblings
        if (elem.level == q.peek().level)
        {
             elem.sibling = q.peek();
             q.peek().sibling = elem;
        }
    }
}

The peek() function gives the next element in the queue without removing it.

I wasn't sure what your question meant exactly. But, I hope this conveys the idea. You could tweak this to suit your needs.

Answer 2

There's actually a rather elegant solution to this. One useful hint is the requirement that this be recursive. When working on a tree, there are several ways to traverse the tree... essentially you can do dfs, or bfs. DFS is naturally recursive and so our solution likely involves dfs (recursive bfs is not at all natural). The next key thing is that in a left-to-right dfs, if we kept track of the last thing seen at each level, we could easily link to our siblings. This (with a bit more simplification) leads to our solution:

void dfs_link(node* n, stack<node*>& st) {
  if (!n) return;
  if (!st.empty()) { st.top()->sibling = n; st.pop(); }
  dfs_link(n->left, st);
  dfs_link(n->right, st);
  st.push(n);
}

void link_tree(node* root) {
  stack<node*> st;
  dfs_link(root, st);
}