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  • 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].

    层序遍历二叉树,依次输出每层最右节点即可,代码如下:

    /**
     * Definition for binary tree
     * 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> re_list = new ArrayList<Integer>();
            List<TreeNode> level_list = new ArrayList<TreeNode>();//层序遍历时存储当前层节点的list
            List<TreeNode> tmp_list = new ArrayList<TreeNode>();//用来存储下一层节点的list
            if(root==null) return re_list;
            level_list.add(root);
            re_list.add(root.val);
            int flag = 1;
            while(flag==1) {
                flag = 0;
                tmp_list.clear();
                for(TreeNode node:level_list) {
                    if(node.left!=null) {
                        flag = 1;//说明下层还存在节点
                        tmp_list.add(node.left);
                    }
                    if(node.right!=null) {
                        flag = 1;//说明下层还存在节点
                        tmp_list.add(node.right);
                    }
                }
                int size = tmp_list.size();
                if(tmp_list.size()!=0) {
                    re_list.add(tmp_list.get(size-1).val);//记录当前层最右边节点
                    level_list.clear();
                    level_list.addAll(tmp_list);//迭代
                }
            }
            return re_list;
        }
    }
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  • 原文地址:https://www.cnblogs.com/mrpod2g/p/4419130.html
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