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         
         int num = 0;
         boolean reverse = false;//a flag
         
         while(!queue.isEmpty()){
             num = queue.size();
             ArrayList<Integer> levelres = new ArrayList<Integer>();
             
             for(int i = 0; i<num; i++){
                 TreeNode node = queue.poll();
                 levelres.add(node.val);
                 
                 if(node.left!=null)
                     queue.add(node.left);
                 if(node.right!=null)
                     queue.add(node.right);
             }
             
             if(reverse){
                 Collections.reverse(levelres);
                 reverse = false;
             }else{
                 reverse = true;
             }
             res.add(levelres);
         }
         
         return res;
     }

 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;
         int currlevel = 1;
         ArrayList<Integer> tmp = new ArrayList<Integer>();
         while(!queue.isEmpty()){
                 TreeNode t = queue.poll();
                 tmp.add(t.val);
                 currlevel--;
                 
                 if(t.left!=null){
                     queue.add(t.left);
                     nextlevel++;
                 }
                 
                 if(t.right!=null){
                     queue.add(t.right);
                     nextlevel++;
                 }
             
             if(currlevel == 0){
                 currlevel = nextlevel;
                 nextlevel = 0;
                 if(reverse){
                     Collections.reverse(tmp);
                     reverse = false;
                 }else{
                     reverse = true;
                 }
                 res.add(tmp);
                 tmp = new ArrayList<Integer>();
             }
         }
         
         return res;
     }
												

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