107. Binary Tree Level Order Traversal II

题目：

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},

    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.

链接： http://leetcode.com/problems/binary-tree-level-order-traversal-ii/

题解：

这几道题都差不多，有关level order的。也可能有很好的方法，但是都没仔细想。基本思路和其他一样，就是BFS

Time Complexity - O(n)， Space Complexity - O(1)。

/**
 * 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<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if(root == null)
            return res;
        ArrayList<Integer> list = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0; 
        
        while(!q.isEmpty()) {
            TreeNode node = q.poll();
            curLevel--;
            list.add(node.val);
            if(node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if(node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if(curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(0, new ArrayList<Integer>(list));
                list.clear();
            }
        }
        
        return res;
    }
}

二刷 :

方法跟一刷一样，就是跟普通level order traversal一样，但是保存结果的时候点到以下顺序。

Java:

Time Complexity - O(n)， Space Complexity - O(1)。

/**
 * 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<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        List<Integer> level = new ArrayList<>();
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            curLevel--;
            level.add(node.val);
            if (node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if (node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(0, new ArrayList<Integer>(level));
                level.clear();
            }
        }
        return res;
    }
}

题外话: 2/14/2016

时间过得真的很快，不抓紧的话根本刷不完。前面一段时间效率极低，每天一两题，也不愿意花时间刷题，反而停留在舒适区，看看书，看看设计。后来返回来刷题发现根本不熟。

最后祝大家情人节快乐。

三刷:

再看了一下discuss，好像用dfs始终比bfs快.

Java:

/**
 * 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<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> res = new ArrayList<>();
        if (root == null) {
            return res;
        }
        List<Integer> level = new ArrayList<>();
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int curLevel = 1, nextLevel = 0;
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            level.add(node.val);
            curLevel--;
            if (node.left != null) {
                q.offer(node.left);
                nextLevel++;
            }
            if (node.right != null) {
                q.offer(node.right);
                nextLevel++;
            }
            if (curLevel == 0) {
                curLevel = nextLevel;
                nextLevel = 0;
                res.add(0, new ArrayList<Integer>(level));
                level.clear();
            }
        }
        return res;
    }
}

Reference:

https://leetcode.com/discuss/5353/there-better-regular-level-order-traversal-reverse-result

https://leetcode.com/discuss/91100/share-java-2ms-recursive-solution

https://leetcode.com/discuss/22538/my-dfs-and-bfs-java-solution

https://leetcode.com/discuss/81189/java-1ms-beats-98%25-using-preorder