1. Binary Tree Level Order Traversal II My Submissions QuestionEditorial Solution
   
   Total Accepted: 79742 Total Submissions: 234887 Difficulty: Easy
   
   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]

]

思路：

参看前一篇

只要用栈存储即可，然后倒入顺序容器中

http://blog.csdn.net/justdoithai/article/details/51346177

/**

 * 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) {

        stack<vector<int>> res;

        vector<vector<int>> res1;

        if(root==NULL)return res1;

        queue<TreeNode*> levelnode;

        vector<int> row;

        levelnode.push(root);

        while(!levelnode.empty()){

            queue<TreeNode*> pre;

            vector<int> tmp;

            while(!levelnode.empty()){

                TreeNode * treetmp=levelnode.front();

                levelnode.pop();

                tmp.push_back(treetmp->val);

                if(treetmp->left!=NULL)pre.push(treetmp->left);

                if(treetmp->right!=NULL)pre.push(treetmp->right);

            }

            res.push(tmp);

            levelnode = pre;

        }

        while(!res.empty()){

            res1.push_back(res.top());

            res.pop();

        }

        return res1;

    }

};