# Leetcode Binary Tree Right Side View Algorithm

I came across a simple, but cool, algorithm on leetcode not too long ago. To practice my writing, I figured I’d try to explain my thought process behind it.

The problem statement is as follows:

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

For example: Given the following binary tree,

    1            <---
/ \
2   3         <---
\   \
5   4       <---


You should return [1, 3, 4].

As per the norm with OJ problems, the most accessible example given does not closely cover a lot of the necessary edge cases to consider when coming up with a sturdy implementation. Let’s try and think about what it means to get the right-most node as we go down the tree. The key realization here is that the right-most node at each level is really just the largest value at each level. This boils down to implementing a modified breadth first search algorithm.

Let’s start with taking a look at a regular BFS algorithm, and then we’ll take a look at how we can modify it. Your typical C++ BFS algorithm might look like this:


struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int val) : val(val), left(NULL), right(NULL) { }
};

vector<int> returnVector;
queue<TreeNode*> q;
if (!root) return returnVector;

// Fill q with root
q.push(root);

while(q.size()) {
returnVector.push_back(q.front()->val);

if (q.front()->left) q.push(q.front()->left);
if (q.front()->right) q.push(q.front()->right);
q.pop();
}

return returnVector;
}



If you haven’t seen this before, here’s a quick and dirty explanation of the code. The idea is to enqueue the first node (root of the tree) and, while the queue’s size is > 0, enqueue any of the node’s children and remove the current node from the queue. This will cause the queue to expand until remaining nodes have no children to add (leaf nodes) and, since we are pushing the children to the back of the queue, we only get to them after we get to all of the nodes at the current level. This gives us a level order traversal.

Given the following tree structure:

      1
/ \
2   3
/   / \
4   5   6


A visual of the queue would look like this:

___       _______       _____       _______       _____       ________
1    =>   1 2 3    =>   2 3    =>   2 3 4    =>   3 4    =>   3 4 5 6  ...
---       -------       -----       -------       -----       --------


From this, we can see, at any given moment, every node in the queue will be at a level equivalent to, or deeper, than that of the front-most. This is important because our modifications must let us store the largest node in a single level in our returnVector before moving onto the first node of the next level. This means we’ll need to know exactly how many nodes are at each level, so we can break appropriately before filling the queue with nodes from another level.

We can do this by maintaining a variable count to hold the number of nodes in the current level. The key is getting count correct from the beginning. This will act as a base that all of the other level counts will be built off of. Observation will reveal that count is equal to the number of nodes in the queue, namely queue.size(), when we enter the while loop for the first time. This provides us a good place to set/reset count every time we perform operations on the current level.

Note how as we iterate through the queue adding nodes from the next level, the queue will contain some nodes from the current level, and some from the next. This is why it is necessary to set count when the queue only contains nodes from the current level to give us an accurate number of nodes we still have to process.

We now need to grab the largest node in the current level. Looking at how we push nodes into the queue (parent->leftChild before parent->rightChild), we can tell that the largest node will always exist at q.back() (also q.at(count-1)).

Now that we have the largest node in the current level, we must store it in our returnVector. With that taken care of, we simply perform a number of steps that, when finished, leave the queue filled only with nodes of the next level. Let’s loop through the count nodes in the queue and, for each one, push its children and pop the current node (standard BFS procedure). Remember to decrement count every time we pop from the queue to maintain an accurate count of nodes we need to process. Once this is completed, the queue will ONLY contain nodes from the new level which gives us a great place to rinse and repeat.

After all is said and done, you might end up with something like the following:


vector<int> rightSideView(TreeNode *root) {
vector<int> ret;
queue<TreeNode*> q;

if(!root) return ret;
q.push(root);

int count = q.size();
while(q.size()) {

ret.push_back(q.back()->val);
count = q.size();

while(count) {
if(q.front()->left)
q.push(q.front()->left);
if(q.front()->right)
q.push(q.front()->right);
q.pop();
count--;
}

}
return ret;
}



Of course, this is not the only way to do it. If you have any improvements to this method or other ideas you’d like to submit, please leave a comment below.

Written on April 26, 2016