【LeetCode】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,#,#,15,7},

    3
   /   9  20
    /     15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

Solution 1: 普通层次遍历，借助queue

/**
 * 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) {      //runtime: 8ms
        vector<vector<int>> vec;
        if(!root)return vec;
        vector<int> v;

        queue<TreeNode*> q1,q2;
        q1.push(root);
        while(!q1.empty()){
            TreeNode* temp = q1.front();
            q1.pop();
            v.push_back(temp->val);
            if (temp->left)
                q2.push(temp->left);
            if (temp->right)
                q2.push(temp->right);

            if(q1.empty()){
                if (!v.empty())
                    vec.push_back(v);
                v.clear();
                swap(q1, q2);
            }
        }
        reverse(vec.begin(),vec.end());　　　　//or vector<vector<int>> ret(vec.rbegin(),vec.rend());return ret;　　　　 return vec;　　　}}

Solution 2: 递归， 待续