LeetCode -- Binary Tree Level Order Traversal II

Question:

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]
]

 

Analysis:

问题描述：给出一棵二叉树，写出它的从底层到最上层的层次遍历。

思路一：一个投机取巧的思路，前面做了从上层到下层的遍历，二者的不同之处仅在于返回的列表的顺序，因此先从上到下遍历，然后将外层列表倒转一下即可。。

/**
 * 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<List<Integer>>();
       List<Integer> l = new ArrayList<Integer>();
       
       if(root == null)
                   return res;
       if(root.left == null && root.right == null) {
                  l.add(root.val);
           res.add(l);
           return res;
       }
       
       TreeNode start; //记录了每一层的开始
       Queue<TreeNode> node = new LinkedList<TreeNode>();
       node.offer(root);
       while(!node.isEmpty()) {
                   TreeNode cur = node.peek();
                   List<Integer> thisLevel = new ArrayList<Integer>();
                   start = null;
                   while(node.peek() != start && !node.isEmpty()) {
                       cur = node.poll();
                       if(start == null) {
                           if(cur.left != null)
                               start = cur.left;
                           else if(cur.right != null)
                               start = cur.right;
                       }
                       
                       thisLevel.add(cur.val);
                       if(cur.left != null)
                           node.offer(cur.left);
                       if(cur.right != null)
                           node.offer(cur.right);
                       
                   }
                   res.add(thisLevel);
       }
       List<List<Integer>> res1 = new ArrayList<List<Integer>>();
          for(int i=0; i<res.size(); i++) {
              res1.add(res.get(res.size() - 1 - i));
          }
       return res1;
    }

}