Binary Tree Zigzag Level Order Traversal

Q:

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]]

Solution:

Reverse the order on each level.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<List<Integer>>();        if (root == null) return result;        boolean reverse = false;        Stack<TreeNode> stack = new Stack<TreeNode>();        stack.push(root);        while (!stack.isEmpty()) {            List<Integer> list = new ArrayList<Integer>();            Stack<TreeNode> level = new Stack<TreeNode>();            while (!stack.isEmpty()) {                TreeNode cur = stack.pop();                list.add(cur.val);                if (reverse) {                    if (cur.right != null)                        level.push(cur.right);                    if (cur.left != null)                        level.push(cur.left);                }                else {                    if (cur.left != null)                        level.push(cur.left);                    if (cur.right != null)                        level.push(cur.right);                }            }            reverse = !reverse;            stack = level;            result.add(list);        }        return result;    }}