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:
Input: root = [1,2,3,null,5,null,4] Output: [1,3,4]
Example 2:
Input: root = [1,null,3] Output: [1,3]
Example 3:
Input: root = [] Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Code
題意就是要每一個 level 最右側的點，因此使用 level order traversal。
/**
* 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) {
vector<int> res;
if(!root) return res;
queue<TreeNode*> q;
q.push(root);
while(!q.empty()) {
int levelSize = q.size();
TreeNode* rightSideNode;
for(int i = 0; i < levelSize; i++) {
auto node = q.front();
rightSideNode = node;
q.pop();
if(node->left) q.push(node->left);
if(node->right) q.push(node->right);
}
res.push_back(rightSideNode->val);
}
return res;
}
};Brilliant ! if(res.size()<level) res.push_back(root->val);
class Solution {
public:
void recursion(TreeNode *root, int level, vector<int> &res)
{
if(root==NULL) return ;
if(res.size()<level) res.push_back(root->val);
recursion(root->right, level+1, res);
recursion(root->left, level+1, res);
}
vector<int> rightSideView(TreeNode *root) {
vector<int> res;
recursion(root, 1, res);
return res;
}
};