[Swift]LeetCode998. 最大二叉树 II | Maximum Binary Tree II

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We are given the root node of a maximum tree: a tree where every node has a value greater than any other value in its subtree.

Just as in the previous problem, the given tree was constructed from an list A (root = Construct(A)) recursively with the following Construct(A) routine:

• If A is empty, return null.
• Otherwise, let A[i] be the largest element of A.  Create a root node with value A[i].
• The left child of root will be Construct([A[0], A[1], ..., A[i-1]])
• The right child of root will be Construct([A[i+1], A[i+2], ..., A[A.length - 1]])
• Return root.

Note that we were not given A directly, only a root node root = Construct(A).

Suppose B is a copy of A with the value val appended to it.  It is guaranteed that B has unique values.

Return Construct(B). 

Example 1:

Input: root = [4,1,3,null,null,2], val = 5
Output: [5,4,null,1,3,null,null,2]
Explanation: A = [1,4,2,3], B = [1,4,2,3,5]

Example 2:

Input: root = [5,2,4,null,1], val = 3
Output: [5,2,4,null,1,null,3]
Explanation: A = [2,1,5,4], B = [2,1,5,4,3]

Example 3:

Input: root = [5,2,3,null,1], val = 4
Output: [5,2,4,null,1,3]
Explanation: A = [2,1,5,3], B = [2,1,5,3,4] 

Note:

1. 1 <= B.length <= 100

---

最大树定义：一个树，其中每个节点的值都大于其子树中的任何其他值。

给出最大树的根节点 root。

就像之前的问题那样，给定的树是从表 A（root = Construct(A)）递归地使用下述 Construct(A) 例程构造的：

• 如果 A 为空，返回 null
• 否则，令 A[i] 作为 A 的最大元素。创建一个值为 A[i] 的根节点 root
• root 的左子树将被构建为 Construct([A[0], A[1], ..., A[i-1]])
• root 的右子树将被构建为 Construct([A[i+1], A[i+2], ..., A[A.length - 1]])
• 返回 root

请注意，我们没有直接给定 A，只有一个根节点 root = Construct(A).

假设 B 是 A 的副本，并附加值 val。保证 B 中的值是不同的。

返回 Construct(B)。 

示例 1：

输入：root = [4,1,3,null,null,2], val = 5
输出：[5,4,null,1,3,null,null,2]
解释：A = [1,4,2,3], B = [1,4,2,3,5]

示例 2：

输入：root = [5,2,4,null,1], val = 3
输出：[5,2,4,null,1,null,3]
解释：A = [2,1,5,4], B = [2,1,5,4,3]

示例 3：

输入：root = [5,2,3,null,1], val = 4
输出：[5,2,4,null,1,3]
解释：A = [2,1,5,3], B = [2,1,5,3,4] 

提示：

1. 1 <= B.length <= 100

---

Runtime: 16 ms

Memory Usage: 18.7 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func insertIntoMaxTree(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        var root = root
        var X:TreeNode? = TreeNode(val)
        if root == nil
        {
            return X
        }
        if root!.val < val
        {
            X!.left = root
            return X
        }        
        dfs(&root, X)
        return root
    }
    
    func dfs(_ root: inout TreeNode?, _ X: TreeNode?)
    {
        if root!.right == nil || root!.right!.val < X!.val
        {
            var Y:TreeNode? = root!.right
            root?.right = X
            X?.left = Y
            return
        }
        dfs(&root!.right, X)
    }
}

---

16ms

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func insertIntoMaxTree(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        return constructMax(root, val)
    }
    
    func constructMax(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        guard let current = root else { return TreeNode(val) }
        if val > current.val {
            let newNode = TreeNode(val)
            newNode.left = current
            return newNode
        }
        current.right = constructMax(current.right, val)
        return current
    }
}