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]
]

解法一：

// Iterative
class Solution {
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        vector<vector<int> > res;
        if (root == NULL) return res;

        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            vector
 
<int> oneLevel;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                oneLevel.push_back(node->val);
                if (node->left) q.push(node->left);
                
 
if (node->right) q.push(node->right);
            }
            res.insert(res.begin(), oneLevel);
        }
        return res;
    }
};

下面我們來看遞迴的解法，核心就在於我們需要一個二維陣列，和一個變數level，當level遞迴到上一層的個數，我們新建一個空層，繼續往裡面加數字，參見程式碼如下：

解法二：

// Recurive
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int> > res;
        levelorder(root, 0, res);
        return vector<vector<int> > (res.rbegin(), res.rend());
    }
    void levelorder(TreeNode *root, int level, vector<vector<int> > &res) {
        if (!root) return;
        if (res.size() == level) res.push_back({});
        res[level].push_back(root->val);
        if (root->left) levelorder(root->left, level + 1, res);
        if (root->right) levelorder(root->right, level + 1, res);
    }
};

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