[Lintcode]Binary Tree Level Order Traversal II 二叉树的层次遍历 II

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).

Example

Given binary tree 

{3,9,20,#,#,15,7}

,

3
/ \
9  20
/  \
15   7

分析：使用队列，对二叉树进行广度遍历。每个元素出队列时，将元素放入栈中。最后按顺序出栈即可

/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/

public class Solution {
/**
* @param root: The root of binary tree.
* @return: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root == null) return res;
LinkedList<TreeNode> in = new LinkedList<TreeNode>();
Stack<TreeNode> out = new Stack<TreeNode>();
TreeNode dummy = new TreeNode(0);
in.add(root);
in.add(dummy);
while(in.size() != 1) {
TreeNode tmp = in.poll();
if(tmp.val == 0) {//change to another layer
TreeNode sep = new TreeNode(0);
out.add(sep);
in.add(tmp);
} else {
out.add(tmp);
if(tmp.right != null) in.add(tmp.right);
if(tmp.left != null) in.add(tmp.left);
}
}
out.add(dummy);

ArrayList<Integer> list = null;
while(!out.isEmpty()) {
TreeNode tmp = out.pop();
if(tmp.val == 0) {
if(list != null) res.add(list);
list = new ArrayList<Integer>();
} else {
list.add(tmp.val);
}
}
if(list != null) res.add(list);
return res;
}

}