Binary Tree Level Order Traversal II

一、题目要求

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.

二、代码实现

vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> result;
queue<TreeNode*> que1,que2;
if(root==NULL)
return result;
//
que1.push(root);
TreeNode* tmp;
vector<int> val;
while(!que1.empty()|| !que2.empty())
{
if(que1.empty())
{
while(!que2.empty())
{
tmp=que2.front();
val.push_back(tmp->val);
if(tmp->left!=NULL)
que1.push(tmp->left);
if(tmp->right!=NULL)
que1.push(tmp->right);
que2.pop();
}
result.insert(result.begin(),val);
val.clear();
}
else
{
while(!que1.empty())
{
tmp=que1.front();
val.push_back(tmp->val);
if(tmp->left!=NULL)
que2.push(tmp->left);
if(tmp->right!=NULL)
que2.push(tmp->right);
que1.pop();
}
result.insert(result.begin(),val);
val.clear();
}
}

return result;
}