binary-tree-level-order-traversal 2

题目描述

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}". 

实现代码：

             /**
 * 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;
        vector<int> level;
        queue<TreeNode*>q;
        int size;
        if(root==NULL) return result;
        q.push(root);
        while(!q.empty())
        {
            level.clear();
            size=q.size();
            for(int i=0;i<size;i++) //此处不能是q.size()；有弹出操作造成队列长度变化；
            {
                TreeNode*p=q.front();
                q.pop();
                level.push_back(p->val);
                if(p->left) q.push(p->left);
                if(p->right) q.push(p->right);
            }
            result.push_back(level);
        }
        reverse(result.begin(),result.end()); //二维vector也可以翻转;
        return result;
    }
};