Maximum Width of Binary Tree
Given a binary tree, write a function to get the maximum width of the given tree. The maximum width of a tree is the maximum width among all levels.
The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null
nodes between the end-nodes are also counted into the length calculation.
It is guaranteed that the answer will in the range of 32-bit signed integer.
Example 1:
Input: 1 / \ 3 2 / \ \ 5 3 9 Output: 4 Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:
Input: 1 / 3 / \ 5 3 Output: 2 Explanation: The maximum width existing in the third level with the length 2 (5,3).
Example 3:
Input: 1 / \ 3 2 / 5 Output: 2 Explanation: The maximum width existing in the second level with the length 2 (3,2).
Example 4:
Input: 1 / \ 3 2 / \ 5 9 / \ 6 7 Output: 8 Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
Constraints:
- The given binary tree will have between
1
and3000
nodes.
/**
* 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:
// long long int max_width;
// unordered_map<long long int,long long int> mp;
public:
// void widthOfBinary(TreeNode* root,long long int depth,long long int pos)
// {
// if(root==NULL)
// return ;
// if(mp.find(depth)==mp.end())
// mp[depth]=pos;
// if(max_width<pos-mp[depth]+1)
// max_width=pos-mp[depth]+1;
// widthOfBinary(root->left,depth+1,2*pos);
// widthOfBinary(root->right,depth+1,2*pos+1);
// }
int widthOfBinaryTree(TreeNode* root)
{
if(root==NULL)
return 0;
int max_width=1;
queue<pair<TreeNode*,int>> q;
q.push({root,0});
while(!q.empty())
{
int l=q.size(),s=0,e=0;
s=q.front().second;
e=q.back().second;
// cout<<e<<" "<<s<<" "<<max_width<<endl;
max_width=max(max_width,e-s+1);
while(l--)
{
auto p=q.front();
q.pop();
long int k=p.second-s;
// cout<<k<<endl;
if(p.first->left)
q.push({p.first->left,2*(k)+1});
if(p.first->right)
q.push({p.first->right,2*(k)+2});
}
}
return max_width;
}
};