LeetCode 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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1  / \ 2   3    /   4    \     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

/** * 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> > iveclist;if (root == NULL)return iveclist;vector<int> rowSize, buf;queue<TreeNode *> TreeFIFO;rowSize.push_back(0);TreeFIFO.push(root);while (!TreeFIFO.empty()) {int n = TreeFIFO.size();for (int i = 0; i < n; i++) {TreeNode *tempNode = TreeFIFO.front();TreeFIFO.pop();buf.push_back(tempNode->val);if (tempNode->left)TreeFIFO.push(tempNode->left);if (tempNode->right)TreeFIFO.push(tempNode->right);}rowSize.push_back(buf.size());}for (int i = rowSize.size() - 2; i >= 0; i--) {vector<int> tempVec;for (int j = rowSize[i]; j < rowSize[i + 1]; j++) {tempVec.push_back(buf[j]);}iveclist.push_back(tempVec);}return iveclist;}};