63. Binary Tree Level Order Traversal II

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;
    }
};