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.

先算下最深，然后再push。。。因为in time的push会出错。。。汗。。。

/** * 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) {        int n=maxDepth(root);        vector< vector<int> > res(n, vector<int>());        helper(root, res, n-1);        return res;    }        void helper(TreeNode* root, vector< vector<int> >& res, int level){        if (root==0)            return;        res[level].push_back(root->val);        helper(root->left, res, level-1);        helper(root->right, res, level-1);    }    int maxDepth(TreeNode* root){        if (root==0)            return 0;        return max(maxDepth(root->left), maxDepth(root->right))+1;    }};