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],

    3
   / \
  9  20
    /  \
   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;
     }