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.
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <---
/ \
2 3 <---
\ \
5 4 <---
在discuss中看到一个很聪明的办法，直接用改动的先根遍历找到结果，改动的地方就是，访问顺序变为，先根后右再左。每次访问时，判断层次和结果大小的关系。因为是先访问右再左，每一层第一个访问的，肯定是最右边的节点。第一个访问的就直接push到结果。
//层次遍历的方法
/**
* 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) {
queue<TreeNode*> qu;
TreeNode* tmp=new TreeNode(-1);
vector<int>res;
if(root==NULL)
return res;
res.push_back(root->val);
qu.push(root->left);
qu.push(root->right);
qu.push(tmp);
int r;
while(!qu.empty()){
TreeNode* t=qu.front();
qu.pop();
if(t==NULL){
continue;
}
if(t==tmp){
// cout<<qu.size()<<' ';
if(qu.empty())
break;
res.push_back(r);
qu.push(tmp);
continue;
}
qu.push(t->left);
qu.push(t->right);
r=t->val;
}
return res;
}
};
//修改的先根遍历的方法
class Solution {
public:
vector<int> rightSideView(TreeNode* root) {
vector<int> res;
modify_preorder(root,1,res);
return res;
}
void modify_preorder(TreeNode* root,int level,vector<int>&res){
if(root==nullptr)
return;
if(res.size()<level)
res.push_back(root->val);
modify_preorder(root->right,level+1,res);
modify_preorder(root->left,level+1,res);
}
};