leetcode 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.

 

这道题目，我想了一天= =

题外话：昨天南京突然降温，超级冷，再加之前天晚上失眠，头脑不太清楚，所以，想这道题想了很久，就是想不对。

想不对的时候乱改一通，又浪费时间又懊丧。

今早真的觉得自己就要做不出来了，后来还是在自己的基础上改对了。

 

我用的方法就是层序遍历+逆序。

层序遍历的时候，我用了一个int值来计每层有多少节点，来判断什么时候得到一个vector。什么时候清空。

思考很重要，耐心很重要。

/**
 * 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>> result;
        if(root==NULL) return result;
        queue<TreeNode*> Tree1;
        int count=1,thiscount=0;
        TreeNode* p;
        Tree1.push(root);
        vector<int> vtree;
        while(!Tree1.empty()){
            p=Tree1.front();
            vtree.push_back(p->val);
            count--;
            Tree1.pop();
            if(p->left!=NULL){
                Tree1.push(p->left);
                thiscount++;
            }
            if(p->right!=NULL){
                Tree1.push(p->right);
                thiscount++;
            }
            if(count==0){
            result.push_back(vtree);
            vtree.clear();
            count=thiscount;
            thiscount=0;
            }
            

        }
        reverse(result.begin(),result.end());
        return result;
    }
};