leetcode 107

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]
]
在102题基础上对输出结果进行反转操作，重新定义了一个vector，将102的结果倒序存入vector中。

代码如下：

/**
 * 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>> lorder;
        vector<vector<int>> lforder;
        if(root == NULL)
        {
            return lorder;
        }
        queue<TreeNode*> q;
        q.push(root);
        vector<int> level_tmp2;
        while(!q.empty())
        {
            level_tmp2.clear();
            queue<TreeNode*> level_tmp1;
            TreeNode* node = q.front();
            int size = q.size();
            
            for(int i = 0; i < size; i++)
            {
                TreeNode* node = q.front();
                q.pop();
                if(node->left)
                {
                    level_tmp1.push(node->left);
                }
                if(node->right)
                {
                    level_tmp1.push(node->right);
                }
                level_tmp2.push_back(node->val);
            }
            while(!level_tmp1.empty())
            {
                q.push(level_tmp1.front());
                level_tmp1.pop();
            }
            lorder.push_back(level_tmp2);
        }
        int n = lorder.size();
        for(int i = n-1; i >= 0; i--)
        {
            lforder.push_back(lorder[i]);
        }//reverse(lorder.begin(), lorder.end());
       return lforder; 
    }
};