Binary Tree Right Side View
LeetCode好久没有总结过了,今天来看一道二叉树Medium
题目
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].
这道题目并不难,找出层次遍历
(BFS)每一层最后一个节点,有趣的地方在于如何优雅准确的一次通过
。
这里用两个队列,并且使用翻转技巧
/**
* 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<int > img;
queue<TreeNode *> q[2];
bool qi=false;
if(root){
q[qi].push(root);
}
while(true){
if(q[qi].size()==0){
break;
}
TreeNode *t;
while(q[qi].size() > 0){
t = q[qi].front();
q[qi].pop();
if(t->left){
q[!qi].push(t->left);
}
if(t->right){
q[!qi].push(t->right);
}
}
img.push_back(t->val);
qi=!qi;
}
return img;
}
};
END geeksword@onestraw.net
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