Leetcode - Binary Tree Level Order Traversal II

My code:

import java.util.ArrayList;
import java.util.Collections;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> levelOrderBottom(TreeNode root) {
        ArrayList> result = new ArrayList>();
        if (root == null)
            return result;
        Queue q = new LinkedList(); 
        q.offer(root);
        while (!q.isEmpty()) {
            int levelSize = q.size();
            ArrayList level = new ArrayList();
            for (int i = 0; i < levelSize; i++) {
                TreeNode temp = q.poll();
                level.add(temp.val);
                if (temp.left != null)
                    q.offer(temp.left);
                if (temp.right != null)
                    q.offer(temp.right);
            }
            result.add(level);
        }
        
        Collections.reverse(result);
        return result;
    }
}

My test result:

Paste_Image.png

主要就是反转下。正好今天学习了
Collections.reverse(); 就用上了，没什么难的。

**
总结： level order of tree. Collections.reverse();
**

Anyway, Good luck, Richardo!

My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> levelOrderBottom(TreeNode root) {
        ArrayList> ret = new ArrayList>();
        if (root == null)
            return ret;
        Queue q = new LinkedList();
        q.offer(root);
        while (!q.isEmpty()) {
            int levelSize = q.size();
            ArrayList group = new ArrayList();
            for (int i = 0; i < levelSize; i++) {
                TreeNode temp = q.poll();
                group.add(temp.val);
                if (temp.left != null)
                    q.offer(temp.left);
                if (temp.right != null)
                    q.offer(temp.right);
            }
            ret.add(group);
        }
        Collections.reverse(ret);
        return ret;
    }
}

void Collections.reverse(xxx);

Anyway, Good luck, Richardo!

iteration 的方法和上面一模一样，就不写了。最后reverse
下面放下 recursion做法，其实也差不多。

My code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> levelOrderBottom(TreeNode root) {
        List> ret = new ArrayList>();
        if (root == null) {
            return ret;
        }
        
        helper(root, 0, ret);
        Collections.reverse(ret);
        return ret;
    }
    
    private void helper(TreeNode root, int level, List> ret) {
        if (root == null) {
            return;
        }
        
        if (level > ret.size() - 1) {
            ret.add(new ArrayList());
        }
        ret.get(level).add(root.val);
        helper(root.left, level + 1, ret);
        helper(root.right, level + 1, ret);
    }
}

就多加一个 Collections.reverse(xxx);
这个函数返回类型是 void

Anyway, Good luck, Richardo! -- 09/06/2016