标签：return ini roo tom example int treenode tree node new
Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ 9 20
/ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
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>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); Queue<TreeNode> q = new LinkedList<>(); if(root == null) { return res; } 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(); level.add(tmp.val); if(tmp.left != null) { q.offer(tmp.left); } if(tmp.right != null) { q.offer(tmp.right); } } res.add(0, 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>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if(root == null) { return res; } levelOrder(root, 0, res); Collections.reverse(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<>()); } res.get(level).add(root.val); levelOrder(root.left, level + 1, res); levelOrder(root.right, level + 1, res); } }
107. Binary Tree Level Order Traversal II - Easy
标签：return ini roo tom example int treenode tree node new
原文地址：https://www.cnblogs.com/fatttcat/p/10193631.html
