标签：== class efi rac amp integer solution from dep
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] ]
Solution 1:
BFS
/** * 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>> arrList = new ArrayList<>(); if (root == null) { return arrList; } Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> list = new ArrayList<>(); for (int i = 0; i < size; i++) { TreeNode cur = queue.poll(); list.add(cur.val); if (cur.left != null) { queue.offer(cur.left); } if (cur.right != null) { queue.offer(cur.right); } } arrList.add(0, list); } return arrList; } }
Solution 2:
/** * 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>> arrList = new ArrayList<>(); helper(arrList, 0, root); return arrList; } private void helper(List<List<Integer>> res, int depth, TreeNode root) { if (root == null) { return; } // backtrack case, resSize >= depth if (depth >= res.size()) { res.add(0, new LinkedList<>()); } res.get(res.size() - 1 - depth).add(root.val); helper(res, depth + 1, root.left); helper(res, depth + 1, root.right); } }
[LC] 107. Binary Tree Level Order Traversal II
标签：== class efi rac amp integer solution from dep
原文地址：https://www.cnblogs.com/xuanlu/p/12151471.html
