leetcode 106: Binary Tree Zigzag Level Order Traversal

Binary
Tree Zigzag Level Order TraversalSep
29 '12

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.

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/

//{1,2,3,4,5}
public class Solution {
public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
// Start typing your Java solution below
// DO NOT write main() function
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
//input check
if(root==null) return res;

Queue<TreeNode> q1 = new LinkedList<TreeNode>();
Queue<TreeNode> q2 = new LinkedList<TreeNode>();

q1.offer(root);
ArrayList<Integer> level = new ArrayList<Integer>();

int i=0;
while( !q1.isEmpty()) {
TreeNode n = q1.poll();
level.add(n.val);

if(n.left!=null) q2.offer(n.left);
if(n.right!=null) q2.offer(n.right);

if(q1.isEmpty()){
if((i&1)==0){
res.add(level);
} else {
Collections.reverse(level);
res.add(level);
}
Queue<TreeNode> temp = q1;
q1 = q2;
q2 = temp;
level = new ArrayList<Integer>();
++i;
}
}
return res;
}
}