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  • [Swift]LeetCode538. 把二叉搜索树转换为累加树 | Convert BST to Greater Tree

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    ➤原文地址:https://www.cnblogs.com/strengthen/p/9817299.html 
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    Given a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus sum of all keys greater than the original key in BST.

    Example:

    Input: The root of a Binary Search Tree like this:
                  5
                /   
               2     13
    
    Output: The root of a Greater Tree like this:
                 18
                /   
              20     13

    给定一个二叉搜索树(Binary Search Tree),把它转换成为累加树(Greater Tree),使得每个节点的值是原来的节点值加上所有大于它的节点值之和。

    例如:

    输入: 二叉搜索树:
                  5
                /   
               2     13
    
    输出: 转换为累加树:
                 18
                /   
              20     13

    80ms
     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public var val: Int
     5  *     public var left: TreeNode?
     6  *     public var right: TreeNode?
     7  *     public init(_ val: Int) {
     8  *         self.val = val
     9  *         self.left = nil
    10  *         self.right = nil
    11  *     }
    12  * }
    13  */
    14 class Solution {
    15     private var sum = 0
    16     func convertBST(_ root: TreeNode?) -> TreeNode? {
    17        // in-order dfs
    18         if root != nil {
    19         convertBST(root?.right)
    20         sum += root!.val
    21         root!.val = sum
    22         convertBST(root?.left)
    23         }
    24         return root
    25     }
    26 }

    84ms

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public var val: Int
     5  *     public var left: TreeNode?
     6  *     public var right: TreeNode?
     7  *     public init(_ val: Int) {
     8  *         self.val = val
     9  *         self.left = nil
    10  *         self.right = nil
    11  *     }
    12  * }
    13  */
    14 class Solution {
    15     
    16     func convertBST(_ root: TreeNode?) -> TreeNode? {
    17         guard let root = root else { return nil }
    18         
    19         traverse(root)
    20         
    21         return root
    22     }
    23     
    24     var sum = 0
    25     
    26     func traverse(_ root: TreeNode?) {
    27         guard let root = root else { return }
    28         
    29         traverse(root.right)
    30         sum += root.val
    31         root.val = sum
    32         traverse(root.left)
    33     }
    34 }

    92ms

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public var val: Int
     5  *     public var left: TreeNode?
     6  *     public var right: TreeNode?
     7  *     public init(_ val: Int) {
     8  *         self.val = val
     9  *         self.left = nil
    10  *         self.right = nil
    11  *     }
    12  * }
    13  */
    14 class Solution {
    15     
    16     var sum = 0
    17     
    18     func convertBST(_ root: TreeNode?) -> TreeNode? {
    19         if root == nil {
    20             return root  
    21         } 
    22         convertBST(root!.right)
    23         root!.val = root!.val + sum
    24         sum = root!.val
    25         convertBST(root!.left)
    26         
    27         return root        
    28     }  
    29 }

    100ms

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public var val: Int
     5  *     public var left: TreeNode?
     6  *     public var right: TreeNode?
     7  *     public init(_ val: Int) {
     8  *         self.val = val
     9  *         self.left = nil
    10  *         self.right = nil
    11  *     }
    12  * }
    13  */
    14 class Solution {
    15      func convertBST(_ root: TreeNode?) -> TreeNode? {
    16         var sum = 0
    17          var stack = Stack()
    18          
    19         var node = root
    20          
    21          while !stack.isEmpty || node != nil {
    22              while node != nil {
    23                  stack.push(node!)
    24                  node = node?.right
    25              }
    26              
    27              node = stack.pop()
    28              sum += node!.val
    29              node!.val = sum
    30              
    31              node = node?.left
    32          }
    33          
    34          return root
    35     }
    36 }
    37 
    38 class Stack {
    39     private var array = [TreeNode]()
    40     
    41     var isEmpty: Bool { return array.isEmpty }
    42     
    43     func push(_ node: TreeNode) {
    44         array.append(node)
    45     }
    46     
    47     func pop() -> TreeNode {
    48         return array.popLast()!
    49     }
    50 }

    120ms

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public var val: Int
     5  *     public var left: TreeNode?
     6  *     public var right: TreeNode?
     7  *     public init(_ val: Int) {
     8  *         self.val = val
     9  *         self.left = nil
    10  *         self.right = nil
    11  *     }
    12  * }
    13  */
    14 class Solution {
    15     func convertBST(_ root: TreeNode?) -> TreeNode? {
    16         _ = convertBST(root, 0)
    17         return root
    18     }
    19 
    20     func convertBST(_ root: TreeNode?, _ num: Int) -> Int {
    21         guard let root = root else {
    22             return num
    23         }
    24         
    25         var num = convertBST(root.right, num)
    26         root.val += num
    27         num = root.val
    28         num = convertBST(root.left, num)
    29         return num
    30     }
    31 }
    
    
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  • 原文地址:https://www.cnblogs.com/strengthen/p/9817299.html
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