103. Binary Tree Zigzag Level Order Traversal -----层序遍历

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,null,null,15,7],

return its zigzag level order traversal as:

[

  [3],

  [20,9],

  [15,7]

]

与上一题类似
不同在于，奇数层 正着存，偶数层，倒着存。

 class Solution {

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

         List<List<Integer>> res = new ArrayList<List<Integer>>();

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

         if(root==null) return res;

         int flag =1;

         queue.add(root);

         while (!queue.isEmpty()) {

             int levlnum = queue.size();

             List<Integer> res_temp = new ArrayList<Integer>();

             for (int i = 0;i<levlnum ;i++ ){ //把每层的左右节点都保存到queue里

                 //并讲当层的值从queue里弹出，加到res_temp 中

             if(queue.peek().left!=null) queue.add(queue.peek().left);

             if(queue.peek().right!=null) queue.add(queue.peek().right);

                 if(flag==1)

                 res_temp.add(queue.poll().val);

                 else

                 res_temp.add(0,queue.poll().val);

             }

             res.add(res_temp);

             flag = 1-flag;

         }

         return res;

     }

巴特西

103. Binary Tree Zigzag Level Order Traversal -----层序遍历

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