[leetcode]103. Binary Tree Zigzag Level Order Traversal二叉树来回遍历

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]
]

题目

思路

1.  use Queue to help BFS

2. once scan current level, make a U-turn, then scan next level

代码

 class Solution {
     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
         List<List<Integer>> result = new ArrayList<>();
         Queue<TreeNode> queue = new LinkedList<>();
         queue.add(root);
         int level = 0;
         // lever order traversal
         while (!queue.isEmpty()) {
             int size = queue.size();
             List<Integer> list = new ArrayList<>();
             for (int i = 0; i < size; i++) {
                 TreeNode node = queue.remove();
                 if (node != null) {
                     list.add(node.val);8
                     queue.add(node.left);
                     queue.add(node.right);
                 }
             }
             if (!list.isEmpty()) {
                 // make a U-turn
                 if (level % 2 == 1) {
                     Collections.reverse(list);
                 }
                 result.add(list);
             }
             level++;
         }
         return result;
     }
 }