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],

]

 

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1

  / \

 2   3

    /

   4

    \

     5

The above binary tree is serialized as  "{1,2,3,#,#,4,#,#,5}".

思路：此题和Binary Tree Level Order Traversal是类似的，就是输出的顺序改变了，在上题的基础之上，我将顺序颠倒输出就可以了。

/**

 * 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> > result;

        if(root==NULL)

            return result;

        queue<TreeNode *> pTree;

        vector<int> data;

        pTree.push(root);

        int count=1;

        while(!pTree.empty())

        {

            data.clear();

            int temp_count=0;

            for(int i=0;i<count;i++)

            {

                TreeNode *tn=pTree.front();

                pTree.pop();

                data.push_back(tn->val);

                if(tn->left)

                {

                    pTree.push(tn->left);

                    temp_count++;

                }

                if(tn->right)

                {

                    pTree.push(tn->right);

                    temp_count++;

                }

            }

            count=temp_count;

            result.push_back(data);

        }

        vector<vector<int> > ret;

        vector<vector<int> >::iterator iter=result.end()-1;

        for(;iter>=result.begin();iter--)

        {

            ret.push_back(*iter);

        }

        return ret;

    }

};