107.Binary Tree Level Order Traversal II
难度Easy
刷题内容
原题连接
- https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
内容描述
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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
˼·1 **- ʱ�临�Ӷ�: O(n)*- �ռ临�Ӷ�: O(n)***
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/**
* Definition for a binary tree node.
* 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) {
if(!root)
{
vector<vector<int> > ret;
return ret;
}
vector<int> v;
v.push_back(root ->val);
vector<vector<int> > temp1 = levelOrderBottom(root ->left);
vector<vector<int> > temp2 = levelOrderBottom(root ->right);
int pos1 = temp1.size() - 1;
int pos2 = temp2.size() - 1;
while(pos1 >= 0 && pos2 >= 0)
{
temp1[pos1].insert(temp1[pos1].end(),temp2[pos2].begin(),temp2[pos2].end());
--pos1;
--pos2;
}
if(pos2 >= 0)
{
auto pos = temp2.begin() + pos2 + 1;
temp1.insert(temp1.begin(),temp2.begin(),pos);
}
temp1.push_back(v);
return temp1;
}
};
˼·2 **- ʱ�临�Ӷ�: O(n)*- �ռ临�Ӷ�: O(n)***
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/**
* Definition for a binary tree node.
* 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> > ret;
if(root != nullptr)
{
vector<TreeNode> v;
v.push_back(*root);
judge(v,ret);
}
return ret;
}
void judge(vector<TreeNode>& v,vector<vector<int> >& ret)
{
vector<TreeNode> v1;
vector<int> v2;
for(int i = 0;i < v.size();i++)
{
v2.push_back(v[i].val);
if(v[i].left != nullptr)
v1.push_back(*(v[i].left));
if(v[i].right != nullptr)
v1.push_back(*(v[i].right));
}
if(v1.size())
judge(v1,ret);
ret.push_back(v2);
}
};
