662. Maximum Width of Binary Tree

Medium

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

解題

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func widthOfBinaryTree(root *TreeNode) int {
    m := make(map[int]int) // 存每個 depth 最左邊的 index
    ans := 0

    var dfs func(*TreeNode, int, int)
    dfs = func(root *TreeNode, depth int, index int) {
        if root == nil { return }

        if _, ok := m[depth]; !ok {
            m[depth] = index
        }

        ans = max(index - m[depth] + 1, ans) 
        // 如果目前 index 到這個 depth 最左邊 index 的寬度大於答案，更新答案

        dfs(root.Left, depth + 1, 2*index)
        dfs(root.Right, depth + 1, 2*index + 1)
    }

    dfs(root, 0, 1)

    return ans
}

 func max(a, b int) int {
     if a > b { return a }
     return b
 }