# Binary Tree Level Order Traversal II ### Source - leetcode: [Binary Tree Level Order Traversal II | LeetCode OJ](https://leetcode.com/problems/binary-tree-level-order-traversal-ii/) - lintcode: [(70) Binary Tree Level Order Traversal II](http://www.lintcode.com/en/problem/binary-tree-level-order-traversal-ii/) ~~~ 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). Example Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its bottom-up level order traversal as: [ [15,7], [9,20], [3] ] ~~~ ### 题解 此题在普通的 [BFS](# "Breadth-First Search, 广度优先搜索") 基础上增加了逆序输出，简单的实现可以使用辅助栈或者最后对结果逆序。 ### Java - Stack ~~~ /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: buttom-up level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (root == null) return result; Stack<ArrayList<Integer>> s = new Stack<ArrayList<Integer>>(); Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { int qLen = q.size(); ArrayList<Integer> aList = new ArrayList<Integer>(); for (int i = 0; i < qLen; i++) { TreeNode node = q.poll(); aList.add(node.val); if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } s.push(aList); } while (!s.empty()) { result.add(s.pop()); } return result; } } ~~~ ### Java - Reverse ~~~ /** * Definition of TreeNode: * public class TreeNode { * public int val; * public TreeNode left, right; * public TreeNode(int val) { * this.val = val; * this.left = this.right = null; * } * } */ public class Solution { /** * @param root: The root of binary tree. * @return: buttom-up level order a list of lists of integer */ public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) { ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>(); if (root == null) return result; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { int qLen = q.size(); ArrayList<Integer> aList = new ArrayList<Integer>(); for (int i = 0; i < qLen; i++) { TreeNode node = q.poll(); aList.add(node.val); if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } result.add(aList); } Collections.reverse(result); return result; } } ~~~ ### 源码分析 Java 中 Queue 是接口，通常可用 LinkedList 实例化。 ### 复杂度分析 时间复杂度为 O(n)O(n)O(n), 使用了队列或者辅助栈作为辅助空间，空间复杂度为 O(n)O(n)O(n).