LeetCode-binary-tree-level-order-traversal-ii

最新推荐文章于 2019-07-22 17:34:41 发布

六月二十七

最新推荐文章于 2019-07-22 17:34:41 发布

阅读量171

点赞数

分类专栏： LeetCode 文章标签： tree-level-order-traversal-ii LeetCode

本文链接：https://blog.youkuaiyun.com/weixin_42146769/article/details/88983493

版权

LeetCode 专栏收录该内容

105 篇文章

订阅专栏

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree{3,9,20,#,#,15,7},

return its bottom-up level order traversal as:

[
  [15,7]
  [9,20],
  [3],
]

confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.ArrayList;
public class Solution {
    public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
        ArrayList<ArrayList<Integer>> arraylist = new ArrayList<>();
        if(root==null) return arraylist;
        ArrayList<TreeNode> nodes = new ArrayList<TreeNode>();
        nodes.add(root);
        while(!nodes.isEmpty()) {
            ArrayList<Integer> list = new ArrayList<Integer>();
            int length = nodes.size();
            for(int i=0;i<length;i++) {
                TreeNode node = nodes.remove(0);
                list.add(node.val);
                if(node.left!=null)
                    nodes.add(node.left);
                if(node.right!=null)
                    nodes.add(node.right);
            }
            arraylist.add(0,list);
        }
        return arraylist;
    }
}