[leetcode]Binary Tree Zigzag Level Order Traversal

Binary Tree Zigzag Level Order Traversal

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]
]

算法思路：

BFS，没什么好讲的，用一个boolean变量标记是正序还是逆序，逆序就用头插法，正序用尾插法。

代码如下：

public class Solution {
     List<List<Integer>> res = new ArrayList<List<Integer>>();
        public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
            if(root == null) return res;
            Queue<TreeNode> q = new LinkedList<TreeNode>();
            Queue<TreeNode> copy = new LinkedList<TreeNode>();
            q.offer(root);
            res.add(new ArrayList<Integer>(Arrays.asList(root.val)));
            boolean tag = false;
            while(!q.isEmpty()){
                TreeNode tem = q.poll();
                if(tem.left != null) copy.offer(tem.left);
                if(tem.right != null) copy.offer(tem.right);
                if(q.isEmpty() && !copy.isEmpty()){
                    List<Integer> list = new LinkedList<Integer>();
                    if(tag){
                        while(!copy.isEmpty()){
                            q.add(copy.peek());
                            list.add(copy.poll().val);
                        }
                    }else{
                        while(!copy.isEmpty()){
                            q.add(copy.peek());
                            list.add(0, copy.poll().val);
                        }
                    }
                    res.add(list);
                    tag = !tag;
                }
            }
            return res;
        }
}

 我用了两个queue来进行层次的区分，网上很多人很巧妙的用了一个null来表示一层的结束。请参考这里。