LeetCode - Binary Tree Zigzag Level Order Traversal

作者：disappearedgod
文章出处：/article/3730169.html
时间：2014-5-26

题目

Binary Tree Zigzag Level Order Traversal

Total Accepted: 11074 Total
Submissions: 41859My Submissions

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:

Given binary tree

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

,

3
/ \
9  20
/  \
15   7

return its zigzag level order traversal as:

[
[3],
[20,9],
[15,7]
]

confused what

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

解法

1.添加奇偶层
多次提交后AC

/**
* 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>> zigzagLevelOrder(TreeNode root) {
//Deque<Integer> stack = new ArrayDeque<Integer>();
ArrayList<ArrayList<Integer>>  retlist = new ArrayList<ArrayList<Integer>> ();

if(root!=null){
Deque<TreeNode> s = new ArrayDeque<TreeNode>();

s.push(root);
boolean stop = false;
boolean odd= true;
while(!stop){
Deque<TreeNode> s2 = new ArrayDeque<TreeNode>();
ArrayList<Integer> list = new ArrayList<Integer>();
while(!s.isEmpty()){

if(odd){
TreeNode t = s.pop();
if(t==null)
continue;
list.add(t.val);
if(t.left!=null)
s2.push(t.left);
if(t.right!=null)
s2.push(t.right);

}else{
TreeNode t = s.pop();
if(t==null)
continue;
list.add(t.val);
if(t.right!=null)
s2.push(t.right);
if(t.left!=null)
s2.push(t.left);

}
}
odd = !odd;
retlist.add(list);
if(s2.isEmpty())
return retlist;
s = s2;
System.gc();

}
}
return retlist;
}
}