[leetcode]103. Binary Tree Zigzag Level Order Traversal@Java解题报告

https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/

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]]

package go.jacob.day806;import java.util.ArrayList;import java.util.LinkedList;import java.util.List;import java.util.Queue;import java.util.Stack;/** *  * @author Jacob * 103. Binary Tree Zigzag Level Order Traversal *  * 两种解法：1.使用两个堆栈；2.使用一个queue，用flag记录当前行打印顺序 */public class Demo2 {/* * 方法一 解法与Binary Tree Level Order Traversal一致 区别是需要用一个flag判断应该正序or逆序打印 */public List<List<Integer>> zigzagLevelOrder(TreeNode root) {List<List<Integer>> res = new ArrayList<List<Integer>>();if (root == null)return res;Queue<TreeNode> queue = new LinkedList<TreeNode>();queue.offer(root);// 用flag控制正序或者逆序打印boolean flag = true;while (!queue.isEmpty()) {int size = queue.size();ArrayList<Integer> list = new ArrayList<Integer>();for (int i = 0; i < size; i++) {TreeNode node = queue.poll();if (flag)list.add(node.val);elselist.add(0, node.val);if (node.left != null)queue.offer(node.left);if (node.right != null)queue.offer(node.right);}flag = !flag;res.add(list);}return res;}/* * 解法二 使用两个堆栈 */public List<List<Integer>> zigzagLevelOrder_2(TreeNode root) {TreeNode c = root;List<List<Integer>> ans = new ArrayList<List<Integer>>();if (c == null)return ans;Stack<TreeNode> s1 = new Stack<TreeNode>();Stack<TreeNode> s2 = new Stack<TreeNode>();s1.push(root);while (!s1.isEmpty() || !s2.isEmpty()) {List<Integer> tmp = new ArrayList<Integer>();while (!s1.isEmpty()) {c = s1.pop();tmp.add(c.val);if (c.left != null)s2.push(c.left);if (c.right != null)s2.push(c.right);}ans.add(tmp);tmp = new ArrayList<Integer>();while (!s2.isEmpty()) {c = s2.pop();tmp.add(c.val);if (c.right != null)s1.push(c.right);if (c.left != null)s1.push(c.left);}if (!tmp.isEmpty())ans.add(tmp);}return ans;}private class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int x) {val = x;}}}