Leetcode-199(Java) Binary Tree Right Side View

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].

 

传送门：https://leetcode.com/problems/binary-tree-right-side-view/

用优先队列层次遍历的方法，每次遍历到一层的最后一个节点并把它保留即可。

 

/**
 * 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 LinkedList<Integer>();
        if(root == null)
            return list;
        //保存每一层的节点，用于取出最后一个节点
        Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
        currentLevel.add(root);
        while(!currentLevel.isEmpty())
        {
            int size = currentLevel.size();
            //lastVal用于保存每层最后一个节点值
            for(int i = 0; i < size - 1; i++)
            {
                TreeNode currentNode = currentLevel.poll();
                if(currentNode.left != null)
                    currentLevel.add(currentNode.left);
                if(currentNode.right != null)
                    currentLevel.add(currentNode.right);
            }
            TreeNode lastNode = currentLevel.poll();
            //把最后一个节点加到队列中
            list.add(lastNode.val);
            if(lastNode.left != null)
                currentLevel.add(lastNode.left);
            if(lastNode.right != null)
                currentLevel.add(lastNode.right);
        }
        return list;
    }
}