# 107. 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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
# Solution
Approach 1: Recursion.
Approach 2: Iteration.
# Code (Python)
Approach 1:
class Solution:
def levelOrderBottom(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
result = []
self._traverse(root, 0, result)
result.reverse()
return result
def _traverse(self, root, level, result):
if not root:
return
if len(result) == level:
result.append([])
result[level].append(root.val)
if root.left:
self._traverse(root.left, level + 1, result)
if root.right:
self._traverse(root.right, level + 1, result)
Approach 2:
# Code (C++)
Approach 1:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
if (root == NULL) return res;
scanLevelOrderBottom(res, root, 1);
return res;
}
private:
void scanLevelOrderBottom(vector<vector<int>>& res, TreeNode *node, int level)
{
if (res.size() < level)
res.insert(res.begin(), vector<int>());
res[res.size()-level].push_back(node->val);
if (node->left)
scanLevelOrderBottom(res, node->left, level+1);
if (node->right)
scanLevelOrderBottom(res, node->right, level+1);
}
};
Approach 2:
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
if (root == NULL) return res;
queue<TreeNode*> nodeQueue;
nodeQueue.push(root);
while (!nodeQueue.empty())
{
vector<int> level;
int nodeQueueCurrSize = nodeQueue.size();
for (int i = 0; i < nodeQueueCurrSize; ++i) // BFS with each level separated.
{
TreeNode *node = nodeQueue.front();
nodeQueue.pop();
level.push_back(node->val);
if (node->left)
nodeQueue.push(node->left);
if (node->right)
nodeQueue.push(node->right);
}
res.insert(res.begin(), level);
}
return res;
}
};