LeetCode之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],
]

与Binary Tree Level Order Traversal一样
只不过在程序最后加上一个reserve函数
/**
* 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>> rst;
BFS(root,0,rst);
reverse(rst.begin(),rst.end());
return rst;
}

void BFS(TreeNode *root, int level, vector<vector<int> > &rst) {
if(!root)return ;
if(level>=rst.size()) rst.push_back(vector<int>());
rst[level].push_back(root->val);
BFS(root->left,level+1,rst);
BFS(root->right,level+1,rst);
}
********************************************************************/

vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int>> rst;
queue<TreeNode*> currentLevel, nextLevel;
currentLevel.push(root);
vector<int> ivec;

if (!root) return rst;
while (!currentLevel.empty()) {
TreeNode *currNode = currentLevel.front();
currentLevel.pop();
if (currNode) {
ivec.push_back(currNode->val);
if(currNode->left) nextLevel.push(currNode->left);
if(currNode->right) nextLevel.push(currNode->right);
}
if (currentLevel.empty()) {
rst.push_back(ivec);
ivec.clear();
swap(currentLevel, nextLevel);
}
}
reverse(rst.begin(),rst.end());
return rst;
}

};