Add Sibling to Binary Tree

Question:
Given a binary tree

struct Node {
Node* leftChild;
Node* rightChild;
Node* nextRight;
}
Populate the nextRight pointers in each node.

The first idea come out should be the BFS tree algorithm, which can traverse the binary tree layer by layer. We only need to connect the sibling for each layer.

<pre> int populateNextRightBFS(Node* root) {
    if (!root)
        return 0;
    queue <Node*> q;
    q.push(root);
    int nextLayer = 0;
    int curLayer = 1;
    Node* prev = NULL;
    while (!queue.empty()) {
        Node* cur = queue.front();
        if (prev)
            prev->sibling = cur;
        if (cur->left) {
            queue.push(cur->left);
            nextLayer++;
        }
        if (cur->right) {
            queue.push(cur->right);
            nextLayer++;
        }
        queue.pop();
        curLayer--;
        prev = cur;
        if (!curLayer) {
            curLayer = nextLayer;
            nextLayer = 0;
            prev = NULL;
        }
    }
    return 0;
}

Above algorithm uses an queue, which can be saved actually. We could use the parent’s sibling to visit children’ siblings. Then we have the following algorithm.

<pre> int populateNextRightExt(Node* root) {
    if (!root)
        return 0;

    if (root->left)
        root->left->sibling= root->right;
    Node* cur = root;
    if (root->right) {
        while (cur->sibling && !cur->sibling->left && !cur->sibling->right)
            cur = cur->sibling;
        if (cur->sibling && cur->sibling->left)
            root->right->sibling = cur->sibling->left;
        else if (cur->sibling && cur->sibling->right)
            root->right->sibling = cur->sibling->left;
        else
            root->right->sibling = NULL;
    }

    populateNextRight(root->right);
    populateNextRight(root->left);

    return 0;
}

P.S If the sibling only points to the node in one-step right, and NULL otherwise, we do not need to search for its sibling far away. The simplified algorithm is as follows.

<pre> int populateNextRight(Node* root) {
    if (!root)
        return 0;

    if (root->left)
        root->left->sibling= root->right;
    if (root->right)
        root->right->sibling = (root->sibling) ? root->sibling->left : NULL;

    populateNextRight(root->right);
    populateNextRight(root->left);

    return 0;
}