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]

]

先输出成arraylist，然后再反转即可。

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {

12         // IMPORTANT: Please reset any member data you declared, as

13         // the same Solution instance will be reused for each test case.

14         ArrayList<TreeNode> queue = new ArrayList<TreeNode>();

15         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

16         if(root == null) return result;

17         queue.add(root);

18         TreeNode nl = new TreeNode(Integer.MIN_VALUE);

19         queue.add(nl);

20         ArrayList<Integer> cur = new ArrayList<Integer>();

21         while(queue.size() != 1){

22             TreeNode rn = queue.remove(0);

23             if(rn.val == Integer.MIN_VALUE){

24                 queue.add(nl);

25                 result.add(cur);

26                 cur = new ArrayList<Integer>();

27             }

28             else{

29                 if(rn.left != null) queue.add(rn.left);

30                 if(rn.right != null) queue.add(rn.right);

31                 cur.add(rn.val);

32             }

33         }

34         result.add(cur);

35         for (int i=0; i<result.size(); i++)

36             if (i % 2 == 1) reverse(result,i);

37         return result;

38     }

39     public void reverse(ArrayList<ArrayList<Integer>> result, int num){

40         int len = result.get(num).size();

41         for(int i = 0; i < len / 2; i ++){

42             int tmp = result.get(num).get(i);

43             result.get(num).set(i, result.get(num).get(len - 1 - i));

44             result.get(num).set(len - 1 - i, tmp);

45         }

46     }

47 }

 第三遍：

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {

12         ArrayList<List<Integer>> result = new ArrayList<List<Integer>>();

13         if(root == null) return result;

14         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();

15         queue.add(root);

16         TreeNode nil = new TreeNode(Integer.MAX_VALUE);

17         queue.add(nil);

18         boolean even = false;

19         ArrayList<Integer> row = new ArrayList<Integer>();

20         while(queue.size() != 1){

21             TreeNode tmp = queue.poll();

22             if(tmp.val != Integer.MAX_VALUE){

23                 row.add(tmp.val);

24                 if(tmp.left != null) queue.add(tmp.left);

25                 if(tmp.right != null) queue.add(tmp.right);

26             } else {

27                 if(even) reverseRow(row);

28                 result.add(row);

29                 row = new ArrayList<Integer>();

30                 queue.add(nil);

31                 even = !even;

32             }

33         }

34         if(even) reverseRow(row);

35         result.add(row);

36         return result;

37     }

38     

39     public void reverseRow(ArrayList<Integer> row){

40         int len = row.size();

41         for(int i = 0; i < len / 2; i ++){

42             int tmp = row.get(i);

43             row.set(i, row.get(len - 1 - i));

44             row.set(len - 1 - i, tmp);

45         }

46     }

47 }