https://leetcode.com/problems/binary-tree-right-side-view/
Given 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:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <---
/
2 3 <---
5 4 <---
代码：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<vector<int> > levelOrder;
vector<int> ans;
if(!root) return ans;
queue<pair<TreeNode*, int>> q;
q.push(make_pair(root, 1));
while(!q.empty()) {
pair<TreeNode*, int> tp = q.front();
q.pop();
if(tp.second > levelOrder.size()) {
vector<int> v;
levelOrder.push_back(v);
}
levelOrder[tp.second - 1].push_back(tp.first->val);
if(tp.first->left)
q.push(make_pair(tp.first->left, tp.second + 1));
if(tp.first->right)
q.push(make_pair(tp.first->right, tp.second + 1));
}
for(int i = 0; i < levelOrder.size(); i ++) {
ans.push_back(levelOrder[i][levelOrder[i].size() - 1]);
}
return ans;
}
};
代码：
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<vector<int> > levelOrder;
vector<int> ans;
if(!root) return ans;
helper(root, levelOrder, 1);
for(int i = 0; i < levelOrder.size(); i ++) {
ans.push_back(levelOrder[i][levelOrder[i].size() - 1]);
}
return ans;
}
void helper(TreeNode* root, vector<vector<int> >& ans, int depth) {
vector<int> v;
if(depth > ans.size()) {
ans.push_back(v);
v.clear();
}
ans[depth - 1].push_back(root -> val);
if(root -> left)
helper(root -> left, ans, depth + 1);
if(root -> right)
helper(root -> right, ans, depth + 1);
}
};
小张写的 code 感觉一晚上都在看层序遍历