[Swift]LeetCode701. 二叉搜索树中的插入操作 | Insert into a Binary Search Tree

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Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Note that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

For example, 

Given the tree:
        4
       / \
      2   7
     / \
    1   3
And the value to insert: 5

You can return this binary search tree:

         4
       /   \
      2     7
     / \   /
    1   3 5

This tree is also valid:

         5
       /   \
      2     7
     / \   
    1   3
         \
          4

---

给定二叉搜索树（BST）的根节点和要插入树中的值，将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 保证原始二叉搜索树中不存在新值。

注意，可能存在多种有效的插入方式，只要树在插入后仍保持为二叉搜索树即可。 你可以返回任意有效的结果。

例如, 

给定二叉搜索树:

        4
       / \
      2   7
     / \
    1   3

和 插入的值: 5

你可以返回这个二叉搜索树:

         4
       /   \
      2     7
     / \   /
    1   3 5

或者这个树也是有效的:

         5
       /   \
      2     7
     / \   
    1   3
         \
          4

---

192ms

/**
 * 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 insertIntoBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        var found = false
        guard var node = root else{ return nil}        
        while found == false{
            if node.val > val{                
                if let l = node.left {
                    node = node.left!
                }else{
                     node.left = TreeNode(val)
                    found = true
                }                
            }else{
                if let r = node.right {
                    node = node.right!
                }else{
                    node.right = TreeNode(val)
                    found = true
                }
            }
        }        
        return root
    }
}

---

Runtime: 196 ms

Memory Usage: 20.2 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 insertIntoBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        if root == nil {return TreeNode(val)}
        if root!.val > val
        {
            root?.left = insertIntoBST(root?.left, val)
        }
        else
        {
            root?.right = insertIntoBST(root?.right, val)
        }
        return root
    }
}

---

200ms

/**
 * 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 insertIntoBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        // let newnode = TreeNode()
        // newnode.val = val
        guard let root = root else { return TreeNode(val) }
       
        //go to left subtree
        if val < root.val {
            if root.left == nil { 
                root.left = TreeNode(val)
            }else{
                insertIntoBST(root.left, val)
            }
            
        }
        //go to right subtree
        if val > root.val {
            if root.right == nil {
                root.right = TreeNode(val)
            }else{
                insertIntoBST(root.right, val)
            }
            
        }
        
        return root
    }
}

---

216ms

/**
 * 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 insertIntoBST(_ root: TreeNode?, _ val: Int) -> TreeNode? {
        guard let root = root else { return TreeNode(val) }
        
        var curr: TreeNode? = root
        var prev: TreeNode = root
        while let node = curr { 
            if node.val > val { 
                curr = node.left
            } else { curr = node.right }
            prev = node
        }
        
        if prev.val > val {
            prev.left = TreeNode(val)
        } else { 
            prev.right = TreeNode(val)
        }
        return root
    }
}