标签：binary tree zigzag level order t
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.
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> list = new ArrayList<List<Integer>>();// 存储结果
if (root == null)
return list;
Queue<TreeNode> q1 = new LinkedList<TreeNode>();// 交替存储相邻两层的结点
Queue<TreeNode> q2 = new LinkedList<TreeNode>();
Queue<TreeNode> temp = null;
List<Integer> subList = null;// 存储一层的结点的值
q1.add(root);
TreeNode top = null;
int flag=0;
while (!q1.isEmpty()) {
subList = new ArrayList<Integer>();
while (!q1.isEmpty())// 循环遍历一层结点并将下一层结点存储到队列中
{
top = q1.peek();
q1.poll();
if (top.left != null)
q2.add(top.left);
if (top.right != null)
q2.add(top.right);
subList.add(top.val);
}
flag++;
if((flag&1)==0)
Collections.reverse(subList);
list.add(subList);
temp = q2;// 交换两个队列的值，使q1一直指向要遍历的那一层
q2 = q1;
q1 = temp;
}
return list;
}
leetcode_103_Binary Tree Zigzag Level Order Traversal
标签：binary tree zigzag level order t
原文地址：http://blog.csdn.net/mnmlist/article/details/44492105
