# 4ms Easy C++ BFS and DFS Solutions

• A simple application of level-order traversal. Just push the last node in each level into the result.

The code is as follows.

``````class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> right;
if (!root) return right;
queue<TreeNode*> toVisit;
toVisit.push(root);
while (!toVisit.empty()) {
TreeNode* rightNode = toVisit.back();
right.push_back(rightNode -> val);
int num = toVisit.size();
for (int i = 0; i < num; i++) {
TreeNode* node = toVisit.front();
toVisit.pop();
if (node -> left) toVisit.push(node -> left);
if (node -> right) toVisit.push(node -> right);
}
}
return right;
}
};
``````

Well, the above code is of BFS. This problem can still be solved using DFS. The code is as follows. Play with it to see how it works :-)

``````class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> right;
if (!root) return right;
rightView(root, right, 0);
}
private:
void rightView(TreeNode* node, vector<int>& right, int level) {
if (!node) return;
if (level == right.size())
right.push_back(node -> val);
rightView(node -> right, right, level + 1);
rightView(node -> left, right, level + 1);
}
};``````

• The DFS one is really smart!

• Hi, Zhuzeng. I also like the DFS solution :-)

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