Swift-二叉树重构

之前的文章中关于二叉树的先序遍历，中序遍历，后序遍历，层次遍历，当然我们可以根据二叉树的先序和中序， 重构二叉树，输出后序，也可以根据二叉树的中序和后序，输出先序~

先序和中序重构

定义二叉树的先序，中序和后序:

var rePreOrder:[String] = ["1","2","4","7","3","5","6","8"]
var reInOrder:[String] = ["4","7","2","1","5","3","8","6"]
var rePostOrder:[String] = ["7","4","2","5","8","6","3","1"]

二叉树重构:

    func reConstructForPost(preOrder: [String],inOrder: [String]) -> TreeNode? {

        if inOrder.count == 0 {
            return nil
        }
        var pre_left:[String] = []
        var pre_right:[String] = []
        var in_left:[String] = []
        var in_right:[String] = []

        let root = TreeNode()
        root.data = preOrder[0] as NSString
        
        var rootIndex = 0
        //获取中序数组中的根节点位置
        for index in 0..

中序和后序重构

    func reConstructForPre(inOrder: [String],postOrder: [String]) -> TreeNode? {
        
        if inOrder.count == 0 {
            return nil
        }

        var in_left:[String] = []
        var in_right:[String] = []
        var post_left:[String] = []
        var post_right:[String] = []
        
        let preRoot = TreeNode()
        preRoot.data = postOrder[postOrder.count-1] as NSString
        
        var rootIndex = 0
        //获取中序数组中的根节点位置
        for index in 0..

代码测试:

var constructTree = ConstructBinaryTree()
var rePreOrder:[String] = ["1","2","4","7","3","5","6","8"]
var reInOrder:[String] = ["4","7","2","1","5","3","8","6"]
var rePostOrder:[String] = ["7","4","2","5","8","6","3","1"]
var constructRoot = constructTree.reConstructForPost(rePreOrder, inOrder: reInOrder)
print("FlyELephant")
print("先序和中序输出后序:")
postOrder(constructRoot)

var preRoot = constructTree.reConstructForPre(reInOrder, postOrder: rePostOrder)
print("\n中序和后序输出先序:")
preOrder(preRoot)
print("")

FlyElephant.png