Leetcode_binary-tree-level-order-traversal-ii

地址：http://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

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.

跟上一题类似，地址：http://oj.leetcode.com/problems/binary-tree-level-order-traversal/

笔者的解题报告：http://blog.csdn.net/flyupliu/article/details/21483317

按照上一题的顺序入vector，最后用STL把vector反序一下即可。

注意对比这两个代码。

参考代码：

/**
 * 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> >res;
        if(!root)
        {
            return res;
        }
        queue<TreeNode*>q;
        q.push(root);
        while(!q.empty())
        {
            int size = q.size();
            vector<int> vec;
            while(size--)
            {
                TreeNode*p = q.front();
                vec.push_back(p->val);
                if(p->left)
                {
                    q.push(p->left);
                }
                if(p->right)
                {
                    q.push(p->right);
                }
                q.pop();
            }
            res.push_back(vec);
        }
        reverse(res.begin(), res.end());
        return res;
    }
};