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;
    }
}