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.
Example:
Input: [1,2,3,null,5,null,4]
Output: [1, 3, 4]
Explanation:
1 <—
/ \
2 3 <—
\ \
5 4 <—
难度：medium
题目：给定二叉树，想像一下你站在树的右边，返回能看到的所有结点，结点从上到下输出。
思路：层次遍历，BFS
Runtime: 1 ms, faster than 79.74% of Java online submissions for Binary Tree Right Side View.Memory Usage: 34.7 MB, less than 100.00% of Java online submissions for Binary Tree Right Side View.
/**
* 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<>();
if (null == root) {
return result;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int qSize = queue.size();
for (int i = 0; i < qSize; i++) {
TreeNode node = queue.poll();
if (node.right != null) {
queue.add(node.right);
}
if (node.left != null) {
queue.add(node.left);
}
if (0 == i) {
result.add(node.val);
}
}
}
return result;
}
}