方法一 BFS
使用二叉树层序遍历的思想 并用size来记录每一层的节点个数,当size==1时就是每一层的最后一个
/**
* 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) {
List<Integer> list = new ArrayList<Integer>();
if(root == null) {
return list;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while(!queue.isEmpty()) {
int size = queue.size();
//保存某一层的节点个数
while(size > 0) {
if(size == 1) {
//说明到某一层的最右边了
list.add(queue.peek().val);
}
TreeNode cur = queue.poll();
size--;
if(cur.left != null) {
queue.offer(cur.left);
}
if(cur.right != null) {
queue.offer(cur.right);
}
}
}
return list;
}
}
方法二 DFS
设置depth是为了只把最右边的节点加入集合 否则就是全树遍历加入集合
/**
* 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 {
List<Integer> list = new ArrayList<>();
public List<Integer> rightSideView(TreeNode root) {
if(root == null) {
return list;
}
dfs(root, 0);
return list;
}
public void dfs(TreeNode root, int depth) {
if(root == null) {
return;
}
if(list.size() == depth) {
list.add(root.val);
}
//每进入一次递归 深度加一
depth++;
dfs(root.right, depth);
dfs(root.left, depth);
}
}