Leetcode Binary Tree Level Order Traversal II
2017-06-28 11:15
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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
return its bottom-up level order traversal as:
代码如下:
/**
* 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>> result;
vector<vector<int>> tmp;
vector<int> data;
if(root == NULL)
return result;
deque<TreeNode*> level;
level.push_back(root);
while(!level.empty())
{
data.clear();
int num = level.size();
for(int i=0;i<num;i++)
{
if(level.front()->left != NULL)
level.push_back(level.front()->left);
if(level.front()->right != NULL)
level.push_back(level.front()->right);
data.push_back(level.front()->val);
level.pop_front();
}
result.push_back(data);
}
for(int i=result.size()-1;i>=0;i--)
tmp.push_back(result[i]);
return tmp;
}
};
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] ]
代码如下:
/**
* 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>> result;
vector<vector<int>> tmp;
vector<int> data;
if(root == NULL)
return result;
deque<TreeNode*> level;
level.push_back(root);
while(!level.empty())
{
data.clear();
int num = level.size();
for(int i=0;i<num;i++)
{
if(level.front()->left != NULL)
level.push_back(level.front()->left);
if(level.front()->right != NULL)
level.push_back(level.front()->right);
data.push_back(level.front()->val);
level.pop_front();
}
result.push_back(data);
}
for(int i=result.size()-1;i>=0;i--)
tmp.push_back(result[i]);
return tmp;
}
};
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