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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
void helper(TreeNode *r, vector<vector<int>> &res, int i){
if(!r) return;
if(i==res.size())
res.push_back(vector<int>());
res[i].push_back(r->val);
helper(r->left, res, i+1);
helper(r->right, res, i+1);
}
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
vector<vector<int>> res;
helper(root, res, 0);
reverse(res.begin(), res.end());
return res;
};
-----------------------------------------------------------------------------------
class Solution {
public:
vector<vector<int>> levelOrderBottom(TreeNode* root) {
if(root==NULL)return {};
queue<TreeNode*> q;
q.push(root);
vector<vector<int>> ans;
while(!q.empty())
{
int s=q.size();
vector<int> v;
while(s--)
{
auto p=q.front();
q.pop();
v.push_back(p->val);
if(p->left)
q.push(p->left);
if(p->right)
q.push(p->right);
}
ans.insert(ans.begin(),v);
}
return ans;
}
};
