力扣面试150题--二叉树的右视图

Day 53

题目描述

思路

采取层序遍历，利用一个high的队列来保存每个节点的高度，highb和y记录上一个节点的高度和节点，在队列中，如果队列中顶部元素的高度大于上一个节点的高度，说明上一个节点就是上一层中最右边的元素，加入数组即可，同时最后需要处理最后一个元素，因为最后一个元素没有能比较的了，需要手动加入数组。

/*** 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> stack=new LinkedList<>();Queue<Integer> high=new LinkedList<>();stack.offer(root);high.offer(0);TreeNode x=root;TreeNode y=root;int highb=0;while(!stack.isEmpty()){if(high.peek()>highb){list.add(y.val);}y=stack.peek();highb=high.peek();x=stack.poll();high.poll();if(x.left!=null){stack.offer(x.left);high.offer(highb+1);}if(x.right!=null){stack.offer(x.right);high.offer(highb+1);}}list.add(x.val);return list;}
}