103.binary-tree-zigzag-level-order-traversal.java

/*
 * @lc app=leetcode id=103 lang=java
 *
 * [103] Binary Tree Zigzag Level Order Traversal
 *
 * https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/
 *
 * algorithms
 * Medium (41.42%)
 * Likes: 984
 * Dislikes: 60
 * Total Accepted: 225.4K
 * Total Submissions: 536.1K
 * Testcase Example: '[3,9,20,null,null,15,7]'
 *
 * 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]
 * ]
 *
 *
 */
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode(int x) { val = x; }
 * }
 */
classSolution {
publicList<List<Integer>> zigzagLevelOrder(TreeNoderoot) {
List<List<Integer>> list = newArrayList<>();
if (root == null) {
returnlist;
 }

Stack<TreeNode> stack1 = newStack<>();
Stack<TreeNode> stack2 = newStack<>();
stack1.push(root);

while (!stack1.empty() || !stack2.empty()) {
List<Integer> level = newArrayList<>();
if (!stack1.empty()) {
while (!stack1.empty()) {
TreeNodenode = stack1.pop();
level.add(node.val);
if (node.left != null) stack2.push(node.left);
if (node.right != null) stack2.push(node.right);
 }
 } else {
while (!stack2.empty()) {
TreeNodenode = stack2.pop();
level.add(node.val);
if (node.right != null) stack1.push(node.right);
if (node.left != null) stack1.push(node.left);
 }
 }
list.add(level);
 }

returnlist;
 }
}