题目描述

• 挺简单的，直接看思路代码吧。
  

思路 & 代码

• 就是一个层序遍历，取每层的最右边结点即可

/*** 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;*     }* }*/// 层级遍历，找出每层最右边值即可
class Solution {List<Integer> ans = new ArrayList<>();LinkedList<TreeNode> queue = new LinkedList<>();public List<Integer> rightSideView(TreeNode root) {if(root == null){return ans;}queue.addLast(root);levelFind(1);return ans;}void levelFind(int nodeNums){if(nodeNums == 0){return;}TreeNode now = null;int nextNums = 0;for(int i = 0; i < nodeNums; i++){now = queue.removeFirst();if(now.left != null){queue.addLast(now.left);nextNums++;}if(now.right != null){queue.addLast(now.right);nextNums++;}}ans.add(now.val);levelFind(nextNums);}
}