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.
For example: Given the following binary tree,
1 <---
/ \
2 3 <---
\ \
5 4 <---
You should return [1, 3, 4]
.
返回一棵二叉树从右边看得到的数。
本质是求二叉树每一层的最右边的一个元素,层序遍历?那样太复杂了。
遍历的时候记录深度,当返回数组的大小等于深度时,说明遍历到这一层时还没有找到最右边的数字,
因为我们是右子树优先遍历的,所以可以把当前值加入到结果中。
/**
* 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:
void dfs(vector<int> &res, TreeNode* root, int dep)
{
if(!root) return ;
if(dep == res.size()) res.push_back(root->val);
if(root->right) dfs(res, root->right, dep + 1);
if(root->left) dfs(res, root->left, dep + 1);
}
vector<int> rightSideView(TreeNode* root) {
vector<int> res, flag;
dfs(res, root, 0);
return res;
}
};