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]
]

1.我的解答（真是烂）

/**
 * 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,result;
        if(root == NULL)
        return result;
        queue<TreeNode*> q1, q2;
        q1.push(root);
        vector<int>v;
        while(!q1.empty()){
            TreeNode* t = q1.front();
            q1.pop();
            v.push_back(t->val);
            if(t->left)
                q2.push(t->left);
            if(t->right)
                q2.push(t->right);
            if(q1.empty()){
                res.push_back(v);
                v.clear();
                while(!q2.empty()){
                q1.push(q2.front());
                q2.pop();
                }
            }
        }
        while(!res.empty()){
            result.push_back(res.back());
            res.pop_back();
        }
        return result;
    }
};

2.别人的代码 （用递归）

First version costs 8ms:

void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level) {
    if (!node) return;
    if (level >= ans.size())
        ans.push_back({});
    ans[level].push_back(node->val);
    levelOrder(ans,node->left,level+1);
    levelOrder(ans,node->right,level+1);
}

vector<vector<int>> levelOrderBottom(TreeNode* root) {
    vector<vector<int>> ans;
    levelOrder(ans,root,0);
    reverse(ans.begin(),ans.end());
    return ans;
}

Second version costs 4ms:

int depth(TreeNode *root) {
    if (!root) return 0;
    return max(depth(root->left),depth(root->right))+1;
}

void levelOrder(vector<vector<int>> &ans, TreeNode *node, int level) {
    if (!node) return;
    ans[level].push_back(node->val);
    levelOrder(ans,node->left,level-1);
    levelOrder(ans,node->right,level-1);
}

vector<vector<int>> levelOrderBottom(TreeNode* root) {
    int d = depth(root);
    vector<vector<int>> ans(d,vector<int> {});
    levelOrder(ans,root,d-1);
    return ans;
}