# 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;
    }
};