2014
11-19

# LeetCode-Binary Tree Level Order Traversal II[二叉树]

### 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).

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]
]


confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ’s Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where ‘#’ signifies a path terminator where no node exists below.

Here’s an example:

   1
/ \
2   3
/
4
\
5


The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

// LeetCode, 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<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int>> 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<vector<int>> &result) {
if (!root) return;

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

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


// LeetCode, 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<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int> > result;
if(root == nullptr) return result;

queue<TreeNode*> current, next;
vector<int> 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;
}
};


Java代码:

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public static  List<List<Integer>> levelOrderBottom(TreeNode root) {
if(root == null) return ans;
Queue<TreeNode> queue = new ArrayDeque<TreeNode>();
TreeNode last = new TreeNode(0);
List<Integer> currentLevel = new ArrayList<Integer>();
while(queue.size() > 1){
TreeNode node = queue.poll();
if(node != last){
if(node.left != null) queue.add(node.left);
if(node.right != null) queue.add(node.right);
}else{
currentLevel.clear();
}