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662.maximum-width-of-binary-tree.cpp
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82 lines (75 loc) · 2.4 KB
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// Tag: Tree, Depth-First Search, Breadth-First Search, Binary Tree
// Time: O(N)
// Space: O(W)
// Ref: -
// Note: -
// 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
//
//
/**
* 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 {
public:
int widthOfBinaryTree(TreeNode* root) {
queue<pair<TreeNode*, long long>> q;
q.emplace(root, 0);
int res = 0;
while (!q.empty()) {
int count = q.size();
int start = q.front().second;
int end = q.back().second;
res = max(res, end - start + 1);
for (int i = 0; i < count; i++) {
auto [cur, index] = q.front();
index -= start;
q.pop();
if (cur->left) {
q.emplace(cur->left, 2 * index + 1);
}
if (cur->right) {
q.emplace(cur->right, 2 * index + 2);
}
}
}
return res;
}
};