Binary Tree Level Order Traversal II

题目：

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.

思想：

在层次遍历的基础上，将最后的结果reverse一下就可以了。

代码：

vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int> >ret;vector<int > temp;if (root == NULL)return ret;queue< TreeNode *> q;TreeNode *p = root;//TreeNode *Tag = new TreeNode();q.push(p);q.push(NULL);while (!q.empty()){p = q.front();while (p != NULL){q.pop();temp.push_back(p->val);if (p->left != NULL){q.push(p->left);}if (p->right != NULL){q.push(p->right);}p = q.front();}q.pop();ret.push_back(temp);temp.clear();if (!q.empty()){q.push(NULL);}}reverse(ret.begin(), ret.end());return ret;}