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 }

 

posted on 2013-11-16 05:48  Step-BY-Step  阅读(199)  评论(0)  编辑  收藏  举报

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