199_binaryTreeRightSideView.md

199. Binary Tree Right Side View 🌟🌟

Given the root of 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.

This question is in continuation with A general approach to level order traversal questions series.

Previous Questions

1. Binary tree level order traversal
2. Binary tree level order traversal - II
3. Binary tree zig-zag level order traversal
4. Average of levels

• Both solutions passed with 0ms runtime.

Recursive Solution

• Simple DFS with right node first.
• if our level is same as result vector's size, then we push the value in the result vector.
• Travel right first and then left.

Code

class Solution {
private:
    void rightSideViewDFS(vector<int>& res, TreeNode* root, int level)
    {
        if (root == NULL) return;

        if (level == res.size()) res.push_back(root->val);

        rightSideViewDFS(res, root->right, level + 1); // first right
        rightSideViewDFS(res, root->left, level + 1); // then left
    }

public:
    vector<int> rightSideView(TreeNode* root)
    {
        vector<int> res;
        rightSideViewDFS(res, root, 0);
        return res;
    }
};

Iterative Solution

• We push last element of queue in the result to get right side view.

Code

class Solution {
public:
    vector<int> rightSideView(TreeNode* root)
    {
        vector<int> res;
        if (root == NULL)
            return res;

        queue<TreeNode*> q;
        q.push(root);

        while (!q.empty()) {
            int sz = q.size();
            for (int i = 0; i < sz; i++) {
                TreeNode* node = q.front(); q.pop();

                if (i == sz - 1) res.push_back(node->val);

                if (node->left != NULL) q.push(node->left);
                if (node->right != NULL) q.push(node->right);
            }
        }
        return res;
    }
};