# Binary Tree Level Order Traversal II Problem

## Description

LeetCode Problem 107.

Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).

Example 1:

``````1
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Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],]
``````

Example 2:

``````1
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Input: root = 
Output: []
``````

Example 3:

``````1
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Input: root = []
Output: []
``````

Constraints:

• The number of nodes in the tree is in the range [0, 2000].
• -1000 <= Node.val <= 1000

## Sample C++ Code

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class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> levelOrder, reverseLevelOrder;
queue<TreeNode*> bfsQ;
queue<int> levelQ;
if (root == NULL)
return levelOrder;
bfsQ.push(root);
levelQ.push(1);

TreeNode* curr;
vector<int> currLevel;
int level;
int lastLevel = 1;
while (!bfsQ.empty()) {
curr = bfsQ.front();
bfsQ.pop();
level = levelQ.front();
levelQ.pop();
if (level > lastLevel) {
levelOrder.push_back(currLevel);
currLevel.clear();
}
if (curr != NULL) {
currLevel.push_back(curr->val);
bfsQ.push(curr->left);
bfsQ.push(curr->right);
levelQ.push(level+1);
levelQ.push(level+1);
}
lastLevel = level;

}
for (vector<vector<int>>::reverse_iterator rvit = levelOrder.rbegin();
rvit != levelOrder.rend(); rvit ++) {
reverseLevelOrder.push_back(*rvit);
}
return reverseLevelOrder;
}
};
``````