BinaryTreeZigzagLevelOrderTraversal.java

/*
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]
]
*/

/**
 * Definition for a binary tree node.
 * 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>> res = new ArrayList<>();
        if (root == null)
            return res;
        Stack<TreeNode> stack1 = new Stack<>();
        Stack<TreeNode> stack2 = new Stack<>();
        stack1.push(root);
        
        while (!stack1.isEmpty() || !stack2.isEmpty()) {
            List<Integer> list = new ArrayList<>();
            while (!stack1.isEmpty()) {
                TreeNode node = stack1.pop();
                if (node.left != null)
                    stack2.push(node.left);
                if (node.right != null)
                    stack2.push(node.right);
                list.add(node.val);
            }
            if (!list.isEmpty())
                res.add(list);
            
            List<Integer> list2 = new ArrayList<>();
            while (!stack2.isEmpty()) {
                TreeNode node = stack2.pop();
                if (node.right != null)
                    stack1.push(node.right);
                if (node.left != null)
                    stack1.push(node.left);
                list2.add(node.val);
            }
            if (!list2.isEmpty())
                res.add(list2);
        }
        return res;
    }
}