Binary Tree Level Order Traversal II
最后更新于：2022-04-02 01:11:14
# 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> levelOrderBottom(TreeNode root) {
ArrayList> result = new ArrayList>();
if (root == null) return result;
Stack> s = new Stack>();
Queue q = new LinkedList();
q.offer(root);
while (!q.isEmpty()) {
int qLen = q.size();
ArrayList aList = new ArrayList();
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> levelOrderBottom(TreeNode root) {
ArrayList> result = new ArrayList>();
if (root == null) return result;
Queue q = new LinkedList();
q.offer(root);
while (!q.isEmpty()) {
int qLen = q.size();
ArrayList aList = new ArrayList();
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).
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