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

I don't need anybody to write code for this.. but just an efficient algorithm.

Thanks

Idea is: 1. Traverse tree with BFS. 2. When you do traversing, you're linking nodes on next level - if node has left and right node, you'll link left to right. If node has next node, then you link rightmost child of current node to leftmost child of next 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);
    }

Connect Nodes at same Level (Level Order Traversal , Write a function to connect all the adjacent nodes at the same level in a binary tree. Example: Input Tree A / \ B C / \ \ D E F Output Tree A--->NULL / \ B-->C--​>  When you do traversing, you're linking nodes on next level - if node has left and right node, you'll link left to right. If node has next node, then you link rightmost child of current node to leftmost child of next node.

Create a vector of linked lists. Do a DFS keeping track of your level, and for each node you find, add it to the linked list of the level. This will run in O(n) which is optimal.

Is this what you want to do?

Link Tree nodes at each level, Idea is: 1. Traverse tree with BFS. 2. When you do traversing, you're linking nodes on next level - if node has left and right node, you'll link left to  Connect Nodes at same Level (Level Order Traversal) Connect nodes at same level using constant extra space; Difference between sums of odd level and even level nodes of a Binary Tree; Connect all nodes to their Left Neighbors in a Binary Tree; Sum of leaf nodes at minimum level; Level with maximum number of nodes

This is not a direct answer to the question and may not be applicable based on your situation. But if you have control over the creation and maintenance of the binary tree, it would probably be more efficient to maintain the links while building/updating the tree.

If you kept both left and right pointers at each level, then it would be "simple" (always easy to say that word when someone else is doing the work) to maintain them. When inserting a new node at a given level, you know its direct siblings from the parent node information. You can adjust the left and right pointers for the three nodes involved (assuming not at the edge of the tree). Likewise, when removing a node, simply update the left and right pointers of the siblings of the node being removed. Change them to point to each other.

How to connect nodes at the same level in a binary tree, Connecting all the nodes at the same level of a binary tree involves linking the right_neighbor pointer of each node (in a tree) to its adjacent node on the same  Create a ArrayList of Linked List Nodes. Do the level order traversal using queue (Breadth First Search). Click here to know about how to level order traversal. For getting all the nodes at each level, before you take out a node from queue, store the size of the queue in a variable, say you call it as levelNodes.

I agree with Thomas Ahle's answer if you want to make all of the row-lists at the same time. It seems that you are only interested in making the list for a one specific row.

Let's say you have a giant tree, but you only want to link the 5th row. There's clearly no point in accessing any node below the 5th row. So just do an early-terminated DFS. Unfortunately, you still have to run through all of the ancestors of every node in the list.

But here's the good news. If you have a perfect binary tree (where every single node branches exactly twice except for the last row) then the first row will have 1 one, the second 2, the third 4, the fourth 8 and the fifth 16. Thus there are more nodes on the last row (16) than all the previous put together (1 + 2 + 4 + 8 = 15), so searching through all of the ancestors is still just O(n), where n is the number of nodes in the row.

The worst case on the other hand would be to have the fifth row consist of a single node with a full binary tree above it. Then you still have to search through all 15 ancestors just to put that one node on the list.

So while this algorithm is really your only choice without modifying your data structure its efficiency relies entirely on how populated the row is compared to higher rows.

Link nodes present in each level of a binary tree in the form of a , Given a binary tree, write an efficient algorithm to link nodes at the same level in the form of a linked list like structure. For example, convert binary tree to the left  Create the tree, a queue to store the nodes and insert the root or starting node in the queue. Create an extra array level of size v (number of vertices) and create a visited array. Run a loop while size of queue is greater than 0.

#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;
        }
    }
}

Populating Next Right Pointers in Each Node, You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition: Level of 1 is 3 Level of 2 is 2 Level of 3 is 1 Level of 4 is 3 Level of 5 is 2 Time Complexity of getLevel() is O(n) where n is the number of nodes in the given Binary Tree. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Tree traversal time complexity, In level order traversal, queue one by one stores nodes of different level. Complexity For converting binary tree into a linked list in O(n) time and O(1) space. My tree_to_linked_lists function does a pre-order traversal, adding nodes to their corresponding linked list (based on depth) in the dictionary lists as the tree is traversed. lists is passed into, and returned from (in its mutated state), each call to tree_to_linked_lists .

27. Trees, A common kind of tree is a binary tree, in which each node contains a Also, all of the nodes that are the same distance from the root comprise a level of the tree. Like linked lists, trees are recursive data structures because they are defined  Level order traversal of binary tree is - 1 2 3 4 5. Time Complexity: O (n^2) in worst case. For a skewed tree, printGivenLevel () takes O (n) time where n is the number of nodes in the skewed tree. So time complexity of printLevelOrder () is O (n) + O (n-1) + O (n-2) + .. + O (1) which is O (n^2).

Special Case: Binary Trees, Each node at depth d has 2 children, so there are 2*2d nodes at d+1, and 2*2d= The number of nodes in the tree is the sum of the number of nodes at each level: We could implements it just like a linked list or a general tree, with pointers: Given a binary tree, print level order traversal in a way that nodes of all levels are printed in separate lines. Note that this is different from simple level order traversal where we need to print all nodes together. Here we need to print nodes of different levels in different lines. A simple solution is to print use the recursive function

Comments
  • It will be costly when updating the tree. Good idea though !
  • This is exactly what I want. But is it possible to make it more efficient? how about a breadth first search? it will give us all the nodes at the same level?
  • A BFS would just require more memory. You'd still have to run through all nodes. Depending on what you want to do with this, you might just keep a pointer in each node, to the next node on its level. This could perhaps be kept updated in logarithmic time.
  • I think BFS might be better, actually. It doesn't require more memory in this case, because it just so happens that the memory needed is exactly the information we're gathering anyway. Still O(n), but you can iterate instead of recursing, so the constant factor should be a bit smaller, if the cost of allocation doesn't dominate.
  • Jason: I'm convinced. BFS will be slightly cheaper and maybe a bit more beautiful
  • OP asked for an efficient algorithm, not code. This answer would benefit from a little explanation of how it worked.