Swift-树的子结构

题目:输入两颗二叉树A和B，判断B是不是A的子结构.
核心代码:

    func hasSubTree(root:TreeNode?,subRoot:TreeNode?) -> Bool {
        var result:Bool = false
        if root != nil && subRoot != nil {
            if root!.data! == subRoot!.data! {
                result = isExistSubTree(root: root, subRoot: subRoot)
            }
            
            if !result {
               result = hasSubTree(root: root?.leftChild, subRoot: subRoot)
            }
            
            if !result {
                result = hasSubTree(root: root?.rightChild, subRoot: subRoot)
            }
        }
        
        return result
    }
    
    func isExistSubTree(root:TreeNode?,subRoot:TreeNode?) -> Bool {
        if subRoot == nil {
            return true
        }
        
        if root == nil {
            return false
        }
        
        if root!.data! != subRoot!.data! {
            return false
        }
        
        return isExistSubTree(root: root?.leftChild, subRoot: subRoot?.leftChild) && isExistSubTree(root: root?.rightChild, subRoot: subRoot?.rightChild)
    }

测试代码:

var node1 = TreeNode(data: "4",leftChild: nil,rightChild: nil)
var node2 = TreeNode(data: "7",leftChild: nil,rightChild: nil)

var node3 = TreeNode(data: "2",leftChild: node1,rightChild: node2)
var node4 = TreeNode(data: "9",leftChild: nil,rightChild: nil)

var node5 = TreeNode(data: "8",leftChild: node4,rightChild: node3)
var node6 = TreeNode(data: "7",leftChild: nil,rightChild: nil)

var rootNode = TreeNode(data: "8",leftChild: node5,rightChild: node6)

var cnode1 = TreeNode(data: "9",leftChild: nil,rightChild: nil)
var cnode2 = TreeNode(data: "2",leftChild: nil,rightChild: nil)
var subRootNode = TreeNode(data: "8",leftChild: cnode1,rightChild: cnode2)

var searchTree:SearchTree = SearchTree()
var result:Bool = searchTree.hasSubTree(root: rootNode, subRoot: subRootNode)

if result {
    print("FlyElephat-包含子🌲")
} else {
    print("FlyElephat-不包含子🌲")
}