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,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its bottom-up level order traversal as:

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

思路：
首先定义一个vector<vector<TreeNode*>> temp来保存二叉树的层次遍历结果，然后将temp中的层次中节点的值倒序赋值给vector<vector<int>> result。

accepted code:

/**
 * 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>> result;
        vector<vector<TreeNode*>> temp;
        if(root==nullptr)
        return result;
        vector<TreeNode*> tk;
        tk.push_back(root);
        temp.push_back(tk);
        int i=0;
        while(temp[i].size()>0)
        {
            tk.clear();
            for(int j=0;j<temp[i].size();j++)
            {
                if(temp[i][j]->left!=nullptr)
                tk.push_back(temp[i][j]->left);
                if(temp[i][j]->right!=nullptr)
                tk.push_back(temp[i][j]->right);
            }
            temp.push_back(tk);
            i++;
        }
        vector<int> num;
        for(int i=temp.size()-2;i>=0;i--)
        {
            for(int j=0;j<temp[i].size();j++)
            {
                num.push_back(temp[i][j]->val);
            }
            result.push_back(num);
            num.clear();
        }
        return result;
    }
};