# O(depth) space recursive C++ solution (not O(1) space but still clean and straightforward)

• Only after I finished the solution I realized that the problem requires only O(1) space. The following solution uses O(depth) space to store the current right node of each level and uses "reversed pre-order" traversal (i.e., parent->right child->left child) for recursion.

``````class Solution {
public:
vector<TreeLinkNode*> rightNodes; // rightNodes[i]: the current right node at level i
void connect(TreeLinkNode *r) {
connectAtLevel(r);
}

void connectAtLevel(TreeLinkNode *r, int level = 0) {
if (!r) return;
if (rightNodes.size() == level) { // node "r" is the rightmost node of this level
r->next = NULL;
rightNodes.push_back(r);
}
else { // use previously stored right node as the ->next of the current node at this level
r->next = rightNodes[level];
rightNodes[level] = r;
}
connectAtLevel(r->right, level+1);
connectAtLevel(r->left, level+1);
}
};
``````

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