## Binary Tree Level Order Traversal II

### 描述

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,#,#,15,7}`,

```
3
/ \
9 20
/ \
15 7
```

return its bottom-up level order traversal as:

```
[
[15,7]
[9,20],
[3],
]
```

### 分析

### 递归版

{% if book.java %}
```java
// Binary Tree Level Order Traversal II
// 递归版，时间复杂度O(n)，空间复杂度O(n)
public class Solution {
public List> levelOrderBottom(TreeNode root) {
List> result = new ArrayList<>();
traverse(root, 1, result);
Collections.reverse(result);
return result;
}

void traverse(TreeNode root, int level,
List> result) {
if (root == null) return;

if (level > result.size())

traverse(root.left, level+1, result);
traverse(root.right, level+1, result);
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Binary Tree Level Order Traversal II
// 递归版，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector > levelOrderBottom(TreeNode *root) {
vector> result;
traverse(root, 1, result);
std::reverse(result.begin(), result.end()); // 比上一题多此一行
return result;
}

void traverse(TreeNode *root, size_t level, vector> &result) {
if (!root) return;

if (level > result.size())
result.push_back(vector());

result[level-1].push_back(root->val);
traverse(root->left, level+1, result);
traverse(root->right, level+1, result);
}
};
```
{% endif %}

### 迭代版

{% if book.java %}
```java
// Binary Tree Level Order Traversal II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public List> levelOrderBottom(TreeNode root) {
List> result = new ArrayList<>();

if(root == null) {
return result;
} else {
current.offer(root);
}

while (!current.isEmpty()) {
ArrayList level = new ArrayList<>(); // elments in one level
while (!current.isEmpty()) {
TreeNode node = current.poll();
}
// swap
Queue tmp = current;
current = next;
next = tmp;
}
Collections.reverse(result);
return result;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Binary Tree Level Order Traversal II
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
vector > levelOrderBottom(TreeNode *root) {
vector > result;
if(root == nullptr) return result;

queue current, next;
vector level; // elments in level level

current.push(root);
while (!current.empty()) {
while (!current.empty()) {
TreeNode* node = current.front();
current.pop();
level.push_back(node->val);
if (node->left != nullptr) next.push(node->left);
if (node->right != nullptr) next.push(node->right);
}
result.push_back(level);
level.clear();
swap(next, current);
}
reverse(result.begin(), result.end()); // 比上一题多此一行
return result;
}
};
```
{% endif %}

### 相关题目

* [Binary Tree Level Order Traversal](binary-tree-tevel-order-traversal.md)
* [Binary Tree Zigzag Level Order Traversal](binary-tree-zigzag-level-order-traversal.md)