662. Maximum Width of Binary Tree

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) {}
 * };
 */
 #define ll long long

class Solution {
public:
    int widthOfBinaryTree(TreeNode* root) {
        if(!root) return 0;
        // 使用 pair 的結構 {目前的節點, 節點的 index}
        queue<pair<TreeNode*, ll>> q;
        q.push({root, 0});

        ll maxWidth = 0;
        while(!q.empty()) {
            ll qSize = q.size();
            // 最左邊 node 的 index
            ll left = q.front().second;
            // 最右邊 node 的 index
            ll right = q.back().second;
            
            // 寬度公式：右邊節點的 index - 左邊節點的 index + 1
            maxWidth = max(maxWidth, right - left + 1);

            // 走訪每一層
            for(int i = 0; i < qSize; i++) {
                auto node = q.front(); q.pop();
                // levelIndex = 目前節點的 index - 最左邊節點的 index
                ll levelIndex = node.second - left;
                if(node.first->left) {
                    // 將左邊節點推到 queue
                    q.push({node.first->left, 2 * levelIndex});
                }
                if(node.first->right) {
                    // 將右邊節點推到 queue
                    q.push({node.first->right, 2 * levelIndex + 1});
                }                
            }
        }
        return maxWidth;
    }
};

• 時間複雜度：
  $O(N)$
• 空間複雜度：
  $O(N)$