leetcode 199. Binary Tree Right Side View 求所能看到的叶子节点 ---------- java

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].

使用队列，层次遍历，每一层的最后一个节点。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> rightSideView(TreeNode root) {
        
        List<Integer> list = new ArrayList();
        if (root == null){
            return list;
        }
        Queue<TreeNode> queue = new LinkedList();
        queue.add(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            for (int i = 0; i < size; i++){
                TreeNode node = queue.poll();
                if (node.left != null){
                    queue.add(node.left);
                }
                if (node.right != null){
                    queue.add(node.right);
                }
                if (i == size - 1){
                    list.add(node.val);
                }
            }
        }
        return list;
    }
}

也可以不使用额外空间。

/**
 * 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> list = new ArrayList();
        getRight(root, list, 0);
        return list;
    }
    
    public void getRight(TreeNode curr, List list, int dep){
        
        if (curr == null)
            return ;
        if (dep == list.size())
            list.add(curr.val);
        getRight(curr.right, list, dep + 1);
        getRight(curr.left, list, dep + 1);
        
    }
}