# leetcode\_662

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.

## Solutions

* Map each node to index when the tree is raveled and is layed out level by level. `root(i), leftchild(2 * i), rightchild(2*i + 1)`
* Then, record the leftmost nodes' index in each level.
* **recursion**

```cpp
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    unordered_map<int, unsigned long> m;
    unsigned long res = 0;
    void dfs(TreeNode * root, int d, unsigned long index) {
        if (!root) return;
        dfs(root->left, d + 1, index * 2);
        dfs(root->right, d + 1, index * 2 + 1);
        if (!m.count(d)) m[d] = index;
        res = max(index - m[d] + 1, res);
    }
    int widthOfBinaryTree(TreeNode* root) {
        dfs(root, 0, 0);
        return res;
    }
};
```

1. **level order traversal**

\`\`\`cpp class Solution { public: int widthOfBinaryTree(TreeNode *root) { queue\<pair\<TreeNode* , unsigned long>> q; if (root) q.emplace(root, 1); unsigned long res = 0, mini, index;

```
    while (!q.empty()) {
        int size = q.size();
        mini = q.front().second;
        while (size--) {
            root = q.front().first;
            index = q.front().second; q.pop();
            if (root->left) q.emplace(root->left, 2 * index);
            if (root->right) q.emplace(root->right, 2 * index + 1);
        }
        res = max(index - mini + 1, res);
    }

    return res;
}
```

}; \`\`