[leetcode]103. Binary Tree Zigzag Level Order Traversal二叉树Z形遍历

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;
    }
}