给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
示例:
输入: [1,2,3,null,5,null,4] 输出: [1, 3, 4] 解释: 1 <--- / \ 2 3 <--- \ \ 5 4 <---
队列实现。
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def rightSideView(self, root: TreeNode) -> List[int]:
if not root:
return []
q = collections.deque([root])
res = []
while q:
size = len(q)
res.append(q[0].val)
for _ in range(size):
node = q.popleft()
if node.right:
q.append(node.right)
if node.left:
q.append(node.left)
return res
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> rightSideView(TreeNode root) {
if (root == null) return Collections.emptyList();
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
List<Integer> res = new ArrayList<>();
while (!q.isEmpty()) {
int size = q.size();
res.add(q.peek().val);
while (size-- > 0) {
TreeNode node = q.poll();
if (node.right != null) q.offer(node.right);
if (node.left != null) q.offer(node.left);
}
}
return res;
}
}