思路一:
对树进行层次遍历(从左往右),保存每层的最后一个节点
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.concurrent.ConcurrentLinkedDeque;
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> list = new ArrayList<>();
//树的层次遍历
ConcurrentLinkedDeque<TreeNode> queue = new ConcurrentLinkedDeque<>();
if (root == null){
return list;
}
queue.add(root);
while (!queue.isEmpty()){
//1.获取队列中的最后一个元素
TreeNode node = queue.peekLast();
list.add(node.val);
//2.从头开始遍历队列中的元素
int count = queue.size();
//2.从头开始遍历队列中的元素
while (count > 0) {
TreeNode node2 = queue.pollFirst();
if (node2.left != null) {
queue.add(node2.left);
}
if (node2.right != null) {
queue.add(node2.right);
}
count--;
}
}
return list;
}
}
思路二:
class Solution {
public List<Integer> rightSideView(TreeNode root) {
List<Integer> list = new ArrayList<>();
if (root == null){
return list;
}
find(root,list,0);
return list;
}
private void find(TreeNode root, List<Integer> list, int depth) {
if (depth == list.size()){
list.add(root.val);
}
if (root.right != null){
find(root.right, list, depth + 1);
}
if (root.left != null){
find(root.left, list, depth + 1);
}
}
}