Binary Tree ZigZag Level Order Traversal leetcode java

题目：

 

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]
]



题解：
这题同样是BFS，用一个flag记录是否需要reverse，如果需要的话就把reverse的结果存储即可。
代码如下：

 1      public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
 2         ArrayList<ArrayList<Integer>> res =  new ArrayList<ArrayList<Integer>>();
 3         
 4          if(root== null)
 5              return res;
 6         
 7         LinkedList<TreeNode> queue =  new LinkedList<TreeNode>();
 8         queue.add(root);
 9         
10          int num = 0;
11          boolean reverse =  false; // a flag
12          
13          while(!queue.isEmpty()){
14             num = queue.size();
15             ArrayList<Integer> levelres =  new ArrayList<Integer>();
16             
17              for( int i = 0; i<num; i++){
18                 TreeNode node = queue.poll();
19                 levelres.add(node.val);
20                 
21                  if(node.left!= null)
22                     queue.add(node.left);
23                  if(node.right!= null)
24                     queue.add(node.right);
25             }
26             
27              if(reverse){
28                 Collections.reverse(levelres);
29                 reverse =  false;
30             } else{
31                 reverse =  true;
32             }
33             res.add(levelres);
34         }
35         
36          return res;
37     }

 1      public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {
 2         ArrayList<ArrayList<Integer>> res =  new ArrayList<ArrayList<Integer>>();
 3          if(root ==  null)
 4              return res;
 5         
 6         LinkedList<TreeNode> queue =  new LinkedList<TreeNode>();
 7         queue.add(root);
 8         Boolean reverse =  false;
 9          int nextlevel = 0;
10          int currlevel = 1;
11         ArrayList<Integer> tmp =  new ArrayList<Integer>();
12          while(!queue.isEmpty()){
13                 TreeNode t = queue.poll();
14                 tmp.add(t.val);
15                 currlevel--;
16                 
17                  if(t.left!= null){
18                     queue.add(t.left);
19                     nextlevel++;
20                 }
21                 
22                  if(t.right!= null){
23                     queue.add(t.right);
24                     nextlevel++;
25                 }
26             
27              if(currlevel == 0){
28                 currlevel = nextlevel;
29                 nextlevel = 0;
30                  if(reverse){
31                     Collections.reverse(tmp);
32                     reverse =  false;
33                 } else{
34                     reverse =  true;
35                 }
36                 res.add(tmp);
37                 tmp =  new ArrayList<Integer>();
38             }
39         }
40         
41          return res;
42     }