Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

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.

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).

Note:

Answer will in the range of 32-bit signed integer.

solution

time complexity : O(n)
space complexity : O(n)

/**
 * 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) {
        if (!root) return 0;
        int ans = 1;
        queue<pair<TreeNode*, int>> q;
        q.push({root, 1});
        while (!q.empty()) {
            int size = q.size();
            ans = max(ans, q.back().second - q.front().second + 1);
            for (int i=0; i<size; ++i) {
                TreeNode* cur = q.front().first;
                unsigned int pos = q.front().second;
                q.pop();
                if (cur->left) q.push({cur->left, pos*2});
                if (cur->right) q.push({cur->right, pos*2+1});
            }
        }
        return ans;
    }
};

設計資料結構，以{node, pos(編號)}為儲存單位，從root開始一層一層往下處理。此處計數方式為：

           1
         /   \
        2     3
       / \   / \  
      4   5 6   7

以此類推。

若當前node的編號為i，則其左子樹的編號為i*2，右子樹為i*2+1。
每次處理一個新的層時，先計算這一層最左邊和最右邊node編號的距離，即為該層的寬度。