Given the root of a binary tree, return 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 rightmost non-null nodes), where the null nodes between the end-nodes that would be present in a complete binary tree extending down to that level are also counted into the length calculation.
It is guaranteed that the answer will in the range of a 32-bit signed integer.
Example 1:
Input: root = [1,3,2,5,3,null,9]
Output: 4
Explanation: The maximum width exists in the third level with length 4 (5,3,null,9).
Example 2:
Input: root = [1,3,2,5,null,null,9,6,null,7]
Output: 7
Explanation: The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).
Example 3:
Input: root = [1,3,2,5]
Output: 2
Explanation: The maximum width exists in the second level with length 2 (3,2).
Constraints:
The number of nodes in the tree is in the range [1, 3000].
-100 <= Node.val <= 100
Intuition
Approach:
1) Do a BFS traversal, as we want the width, builing the intuition
2) Now, we can index the tree nodes like
0
/ \
1 2
.. son
if index = i
Left = 2*i+1
right = 2*i+2
in 0 based indexing
now we can maintain a queue to track the indices
Now to avoid overflow, We do a trick
For skewed tree, i - 2i - 4i .. can cause overflow
So mmin trick
class Solution {
public:
int widthOfBinaryTree(TreeNode* root) {
queue<pair<TreeNode*, long long int>> q;
q.push({root, 0});
long long int len=0;
while(!q.empty()){
int size = q.size();
int mmin = q.front().second;
long long int left, right;
for(int i=0; i<size; i++){
TreeNode* temp = q.front().first;
long long int index = q.front().second;
if(i==0)
left = index;
if(i==size-1)
right = index;
if(temp->left){
long long int node_idx = (2*index+1) - mmin;
q.push({temp->left, node_idx});
}
if(temp->right){
long long int node_idx = (2*index+2) - mmin;
q.push({temp->right, node_idx});
}
q.pop();
}
len = max(len, right-left+1);
}
return len;
}
};
