# 0662. Maximum Width of Binary Tree

https://leetcode.com/problems/maximum-width-of-binary-tree

## Description

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 are also counted into the length calculation.

It is **guaranteed** that the answer will in the range of **32-bit** signed integer.

**Example 1:**

![](https://assets.leetcode.com/uploads/2021/05/03/width1-tree.jpg)

```
**Input:** root = [1,3,2,5,3,null,9]
**Output:** 4
**Explanation:** The maximum width existing in the third level with the length 4 (5,3,null,9).
```

**Example 2:**

![](https://assets.leetcode.com/uploads/2021/05/03/width2-tree.jpg)

```
**Input:** root = [1,3,null,5,3]
**Output:** 2
**Explanation:** The maximum width existing in the third level with the length 2 (5,3).
```

**Example 3:**

![](https://assets.leetcode.com/uploads/2021/05/03/width3-tree.jpg)

```
**Input:** root = [1,3,2,5]
**Output:** 2
**Explanation:** The maximum width existing in the second level with the length 2 (3,2).
```

**Example 4:**

![](https://assets.leetcode.com/uploads/2021/05/03/width4-tree.jpg)

```
**Input:** root = [1,3,2,5,null,null,9,6,null,null,7]
**Output:** 8
**Explanation:** The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
```

**Constraints:**

* The number of nodes in the tree is in the range `[1, 3000]`.
* `-100 <= Node.val <= 100`

## ac1: DFS

Will overflow.

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;

        Map<Integer, int[]> map = new HashMap<>();
        helper(root, 0, 0, map);

        int max = Integer.MIN_VALUE;
        for (int[] v : map.values()) {
            max = Math.max(max, v[1] - v[0] + 1);
        }

        return max;
    }

    public void helper(TreeNode node, int pos, int level, Map<Integer, int[]> map) {
        if (node == null) return;
        if (!map.containsKey(level)) {
            map.put(level, new int[]{pos, pos});
        }

        map.get(level)[0] = Math.min(map.get(level)[0], pos);
        map.get(level)[1] = Math.max(map.get(level)[1], pos);

        helper(node.left, pos * 2, level+1, map);
        helper(node.right, pos * 2 + 1, level+1, map);
    }
}

/*
1) Map, record each level's start and end position; 2) careful: overflow, just initialize the value to the first met position
*/
```

## ac2: BFS

[https://leetcode.com/problems/maximum-width-of-binary-tree/discuss/106653/Java-One-Queue-Solution-with-HashMap](https://leetcode.com/problems/maximum-width-of-binary-tree/discuss/106653/Java-One-Queue-Solution-with-HashMap)

More efficient.

```java
class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        Map<TreeNode, Integer> nodeToIndex = new HashMap<>();
        Queue<TreeNode> q = new LinkedList<>();
        int max = Integer.MIN_VALUE;

        q.offer(root);
        nodeToIndex.put(root, 1);

        while(!q.isEmpty()) {
            int start = Integer.MAX_VALUE;
            int end = 0;
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode h = q.poll();
                int index = nodeToIndex.get(h);
                if (i == 0) {
                    start = index;
                }
                if (i == size - 1) {
                    end = index;
                }

                if (h.left != null) {
                    q.offer(h.left);
                    nodeToIndex.put(h.left, 2 * index);
                }
                if (h.right != null) {
                    q.offer(h.right);
                    nodeToIndex.put(h.right, 2 * index + 1);
                }
            }
            max = Math.max(max, end - start + 1);
        }

        return max;
    }
}
```
