Link Tree Nodes at Each Level

Question

Given the binary tree, how would you join nodes at each level from left to right. Say there are 5 nodes in the third level, connect all of them from left to right.

I do not need anyone to write code for this, but just an efficient algorithm.

thank

Answer 1

Create a vector of linked lists. Make DFS, tracking your level, and for each node found, add it to the linked level list. This will work in O (n), which is optimal.

Is that what you want to do?

Answer 2

:
1. BFS.
2. , - node node, . node node, node node.

    public void BreadthFirstSearch(Action<Node> currentNodeAction)
    {
        Queue<Node> q = new Queue<Node>();
        q.Enqueue(root);

        while (q.Count != 0)
        {
            Node current = q.Dequeue();

            if (currentNodeAction != null)
                currentNodeAction(current);

            if (current.left != null) q.Enqueue(current.left);
            if (current.right != null) q.Enqueue(current.right);
        }
    }

    private void Linker(Node node)
    {
        Link(node.left, node.right);

        if (node.next != null)
            Link(node.right ?? node.left, node.next.left ?? node.next.right);
    }

    private void Link(Node node1, Node node2)
    {
        if (node1 != null && node2 != null)
            node1.next = node2;
    }

    public void LinkSameLevel()
    {
        BreadthFirstSearch(Linker);
    }

Answer 3

. , , / .

, , "" ( , - ), . node node . ( ). , node node, . , .

Answer 4

, . , .

, , 5- . node . DFS . , node .

. ( node , ), 1, , , 16. , (16), , (1 + 2 + 4 + 8 = 15), - O (n), n - .

node . 15 , node .

, , , .

Answer 5

#include <queue>

struct Node {
    Node *left;
    Node *right;
    Node *next;
};

/** Link all nodes of the same level in a binary tree. */
void link_level_nodes(Node *pRoot)
{
    queue<Node*> q;
    Node *prev;     // Pointer to the revious node of the current level
    Node *node;
    int cnt;        // Count of the nodes in the current level
    int cntnext;    // Count of the nodes in the next level

    if(NULL == pRoot)
        return;

    q.push(pRoot);
    cnt = 1;
    cntnext = 0;
    prev = NULL;

    while (!q.empty()) {
        node = q.front();
        q.pop();

        /* Add the left and the right nodes of the current node to the queue
           and increment the counter of nodes at the next level. 
        */
        if (node->left){
            q.push(node->left);
            cntnext++;
        }
        if (node->right){
            q.push(node->right);
            cntnext++;
        }

        /* Link the previous node of the current level to this node */
        if (prev)
            prev->next = node;

        /* Se the previous node to the current */
        prev = node;

        cnt--;
        if (0 == cnt) { // if this is the last node of the current level
            cnt = cntnext;
            cntnext = 0;
            prev = NULL;
        }
    }
}

Answer 6

, , , .

, node. , 0.

public Node(int d)
{
head=this;
data=d;
left=null;
right=null;
level=0;
}

, , , . , , Vector of Nodes.

Answer 7

. , , -

1) BFS.
  , . , node node . node, dequeued node, , , .
O (n).

2) , . .
O (n).

3) (2), , , , , , + 1 . O (n ^ 2).

Answer 8

    private class Node
    {
        public readonly Node Left;
        public readonly Node Right;
        public Node Link { get; private set; }

        public void Run()
        {
            LinkNext = null;
        }

        private Node LinkNext
        {
            get
            {
                return Link == null ? null : (Link.Left ?? Link.Right ?? Link.LinkNext);
            }
            set
            {
                Link = value;
                if (Right != null)
                    Right.LinkNext = LinkNext;
                if (Left != null)
                    Left.LinkNext = Right ?? LinkNext;
            }
        }
    }