leetcode_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]]

第一感觉是要用BFS，每一层记一个level，然后下一层这个level+1.但好像迭代的方法用DFS实现是一样的。

代码：

/** * 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>> outv;        void dfs(TreeNode* root, int level){        if(root == NULL) return;        if(level == outv.size()){            outv.push_back(vector<int>());        }        outv[level].push_back(root->val);        dfs(root->left, level+1);        dfs(root->right, level+1);    }     vector<vector<int>> levelOrderBottom(TreeNode* root) {        dfs(root, 0);        return vector<vector<int> > (outv.rbegin(), outv.rend());    }};