# Binary Tree Zigzag Level Order Traversal

### 描述​

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]]

### 递归版​

// Binary Tree Zigzag Level Order Traversal// 递归版，时间复杂度O(n)，空间复杂度O(n)public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        traverse(root, 1, result, true);        return result;    }    private static void traverse(TreeNode root, int level, List<List<Integer>> result,                                 boolean left_to_right) {        if (root == null) return;        if (level > result.size())            result.add(new ArrayList<>());        if (left_to_right)            result.get(level-1).add(root.val);        else            result.get(level-1).add(0, root.val);        traverse(root.left, level+1, result, !left_to_right);        traverse(root.right, level+1, result, !left_to_right);    }}

### 迭代版​

// Binary Tree Zigzag Level Order Traversal// 广度优先遍历，用一个bool记录是从左到右还是从右到左，每一层结束就翻转一下。// 迭代版，时间复杂度O(n)，空间复杂度O(n)public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        Queue<TreeNode> current = new LinkedList<>();        Queue<TreeNode> next = new LinkedList<>();        boolean left_to_right = true;        if(root == null) {            return result;        } else {            current.offer(root);        }        while (!current.isEmpty()) {            ArrayList<Integer> level = new ArrayList<>(); // elments in one level            while (!current.isEmpty()) {                TreeNode node = current.poll();                level.add(node.val);                if (node.left != null) next.offer(node.left);                if (node.right != null) next.offer(node.right);            }            if (!left_to_right) Collections.reverse(level);            result.add(level);            left_to_right = !left_to_right;            // swap            Queue<TreeNode> tmp = current;            current = next;            next = tmp;        }        return result;    }}