題目描述:
給定一棵二叉樹,想象自己站在它的右側,按照從頂部到底部的順序,傳回從右側所能看到的節點值。
示例:
輸入: [1,2,3,null,5,null,4]
輸出: [1, 3, 4]
解釋:
1 <---
/ \
2 3 <---
\ \
5 4 <---
思路:得到題目的右視圖,即要求得到每一層最後一個節點。按照層序周遊,将每一層的最後一個節點添加到res中即可。
/**
* 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> res;
vector<int> rightSideView(TreeNode* root) {
if(root==NULL)
return {};
queue<TreeNode*> que;
que.push(root);
while(!que.empty()){
int size=que.size();
TreeNode* now;
while(size--){
now=que.front();
que.pop();
if(now->left)
que.push(now->left);
if(now->right)
que.push(now->right);
}
res.push_back(now->val);
}
return res;
}
};

/**
* 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> res;
void traverse(TreeNode* root,int level){
if(root==NULL)
return ;
if(res.size()==level){
res.push_back(root->val);
}
traverse(root->right,level+1);
traverse(root->left,level+1);
}
vector<int> rightSideView(TreeNode* root) {
if(root==NULL)
return res;
traverse(root,0);
return res;
}
};