[LeetCode] Binary Tree Right Side View

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);
    }
};