标签:between tree queue lin == rev init bin style
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] ]
M1: BFS
time: O(n), space: O(2^height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); Queue<TreeNode> q = new LinkedList<>(); if(root == null) { return res; } boolean reverse = false; q.offer(root); while(!q.isEmpty()) { List<Integer> level = new ArrayList<>(); int size = q.size(); for(int i = 0; i < size; i++) { TreeNode tmp = q.poll(); if(reverse) { level.add(0, tmp.val); } else { level.add(tmp.val); } if(tmp.left != null) { q.offer(tmp.left); } if(tmp.right != null) { q.offer(tmp.right); } } reverse = !reverse; res.add(level); } return res; } }
M2: recursion
time: O(n), space: O(height)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<List<Integer>> zigzagLevelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if(root == null) { return res; } levelOrder(root, 0, res); return res; } public void levelOrder(TreeNode root, int level, List<List<Integer>> res) { if(root == null) { return; } if(res.size() == level) { res.add(new ArrayList<>()); } if(level % 2 == 0) { res.get(level).add(root.val); } else { res.get(level).add(0, root.val); } levelOrder(root.left, level + 1, res); levelOrder(root.right, level + 1, res); } }
103. Binary Tree Zigzag Level Order Traversal - Medium
标签:between tree queue lin == rev init bin style
原文地址:https://www.cnblogs.com/fatttcat/p/10193632.html