leetCode 107. Binary Tree Level Order Traversal II 二叉樹層次遍歷反轉

107. 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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

解題思路：

此題與Binary Tree Level Order Traversal相似，只是最后的結果有一個反轉。

參考 http://qiaopeng688.blog.51cto.com/3572484/1834819

代碼如下：

/**
 * 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;
        queue<TreeNode *> current,next;
        vector<int> level;
        if(NULL == root)
            return result;
        current.push(root);
        
        while(current.size())
        {
            while(current.size())
            {
                TreeNode *p;
                p = current.front();
                current.pop();
                level.push_back(p->val);
                if(p->left)
                    next.push(p->left);
                if(p->right)
                    next.push(p->right);
            }
            result.push_back(level);
            level.clear();
            swap(current,next);
        }
        reverse(result.begin(),result.end());
        //相對與Binary Tree Level Order Traversal只加了這一句。reverse()，反轉
        return result;
    }
};