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],
]

题意：给一棵二叉树，返回层次遍历后，从叶节点到根回溯的结果。

解题思路：对二叉树进行层次遍历，用count来记录每层的结点数，用队列que实现，最后将层次遍历的结果反序则可。代码应该不难看懂。

/**
 * Definition for binary tree
 * 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> > ans;
        vector<vector<int> > result;
        TreeNode* cur;
        int count = 0;
        queue<TreeNode*> que;
        if(root != NULL) {
            //先处理根结点
            vector<int> tmp;     
            tmp.push_back(root->val);
            ans.push_back(tmp);
            if(root->left != NULL) { count++; que.push(root->left); }
            if(root->right != NULL){ count++; que.push(root->right); }
            while(!que.empty())
            {
                int k = count;
                vector<int> tt;
                count = 0;
                for(int i=0;i<k; i++)
                {
                    cur = que.front();
                    que.pop();
                    tt.push_back(cur->val);
                    if(cur->left != NULL) { count++; que.push(cur->left);  }
                    if(cur->right != NULL){ count++; que.push(cur->right);  }
                }
                ans.push_back(tt);
            }
        }
        if(ans.size() > 0)
        {
            for(int j = ans.size()-1; j >= 0; j--)
                result.push_back(ans[j]);
        }
        return result;
    }
};