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.
【题意】

    二叉树的层次遍历，但是需要逆序输出结果。

【分析】

     二叉树的层次遍历，利用两个队列，从根节点开始，放到第一个队列中，下一层次的节点放到另一个队列中，以此类推！（注意对空树的处理）

【实现】

     /* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> totalList = new ArrayList<ArrayList<Integer>>();
if (root == null) return totalList;

ArrayList<TreeNode> list1 = new ArrayList<TreeNode>();
ArrayList<TreeNode> list2 = new ArrayList<TreeNode>();

list1.add(root);

while (list1.isEmpty() == false) {
ArrayList<Integer> valueList = new ArrayList<Integer>();
for (int i = 0; i < list1.size(); i++) {
TreeNode node = list1.get(i);
valueList.add(node.val);
if (node.left != null) list2.add(node.left);
if (node.right != null) list2.add(node.right);
}
totalList.add(valueList);
ArrayList<TreeNode> temp = list1;
list1 = list2;
list2 = temp;
list2.clear();
}

reverseList(totalList);
return totalList;
}

public void reverseList(ArrayList<ArrayList<Integer>> totalList) {
if (totalList == null || totalList.size() == 1) return ;
int i = 0;
int j = totalList.size() - 1;
while (i < j) {
ArrayList<Integer> temp = totalList.get(i);
totalList.set(i, totalList.get(j));
totalList.set(j, temp);
i++;
j--;
}
}
}