199. Binary Tree Right Side View

Posted on Edited on In

199. Binary Tree Right Side View

Description

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.

Example 1:

1
2
Input: root = [1,2,3,null,5,null,4]
Output: [1,3,4]

Example 2:

1
2
Input: root = [1,null,3]
Output: [1,3]

Example 3:

1
2
Input: root = []
Output: []

Constraints:

  • The number of nodes in the tree is in the range [0, 100].
  • -100 <= Node.val <= 100

Hints/Notes

Solution

Language: C++

Recursive solution:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> res;
    vector<int> rightSideView(TreeNode* root) {
        dfs(root, 0);
        return res;
    }

    void dfs(TreeNode* root, int depth) {
        if (!root) {
            return;
        }
        if (res.size() == depth) {
            res.push_back(root->val);
        }
        if (root->right) {
            dfs(root->right, depth + 1);
        }
        if (root->left) {
            dfs(root->left, depth + 1);
        }
    }
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        if (!root) {
            return {};
        }
        queue<TreeNode*> q;
        q.push(root);
        vector<int> res;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                TreeNode* node = q.front();
                q.pop();
                if (i == 0) {
                    res.push_back(node->val);
                }
                if (node->right) {
                    q.push(node->right);
                }
                if (node->left) {
                    q.push(node->left);
                }
            }
        }
        return res;
    }
};