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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].
Credits:
Special thanks to @amrsaqr for adding this problem and creating all test cases.
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思路：
层次遍历。
c++ code:
/**
* 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> ret;
if (!root) return ret;
queue<TreeNode*> que;
que.push(root);
while (!que.empty()) {
int size = que.size();
for (int i = 0; i < size; i++) {
TreeNode *tmp = que.front();
que.pop();
if (i == 0) ret.push_back(tmp->val);
if (tmp->right) que.push(tmp->right);
if (tmp->left) que.push(tmp->left);
}
}
return ret;
}
};java code:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> result = new ArrayList<Integer>();
if(root == null) return result;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
for(int i=0;i<size;i++) {
TreeNode tmp = queue.poll();
if(i==0) result.add(tmp.val);
if(tmp.right != null) queue.offer(tmp.right);
if(tmp.left != null) queue.offer(tmp.left);
}
}
return result;
}
}LeetCode:Binary Tree Right Side View
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原文地址：http://blog.csdn.net/itismelzp/article/details/51636280
