虽然我在工作中从没有用到过二叉树这个结构,但是觉得还是挺重要的。emmm,就是这样。
二叉树:二叉树的每个节点最多有两个子节点,一般称为左子节点和右子节点,二叉树有左子树和右子树之分。
节点实现
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
}
}
二叉树是由递归实现的,所以理论上所有的二叉树问题都可以通过递归的方法来实现。
计算二叉树最大深度
func maxDepth(_ root: TreeNode?) -> Int {
guard let root = root else { return 0 }
return max(maxDepth(root.left), maxDepth(root.right)) + 1
}
判断一棵二叉树是否二叉查找树(二叉查找树:所有左子树节点的值都小于根节点的值,所有右子树节点的值都大于根节点的值)
func isValidBST(_ root: TreeNode?) -> Bool{
return helper(root, nil, nil)
}
func helper(_ node: TreeNode?, _ min: Int?, _ max:Int?) -> Bool{
guard let node = node else {
return true
}
if let min = min, node.val <= min {
return false
}
if let max = max, node.val >= max{
return false
}
return helper(node.left, min, node.val) && helper(node.right, node.val, max)
}
翻转一颗二叉树
func invertTree(_ root: TreeNode?) -> TreeNode? {
if root == nil {
return nil
}
let left = invertTree(root?.left)
let right = invertTree(root?.right)
root?.left = right
root?.right = left
return root
}
两棵二叉树求和
func MergeTree(_ t1: TreeNode?, _ t2: TreeNode?) -> TreeNode? {
if (t1 == nil || t2 == nil) {return t1 == nil ? t2 : t1}
let val = t1!.val + t2!.val
let root = TreeNode(val)
root.left = MergeTree(t1?.left, t2?.left)
root.right = MergeTree(t1?.right, t2?.right)
return root
}
递归遍历二叉树
func printTree(treeNode:TreeNode?) -> Int {
if treeNode == nil {
return 0
}
print(treeNode?.val as Any)
printTree(treeNode: treeNode?.left)
printTree(treeNode: treeNode?.right)
return 1
}
非递归遍历二叉树(用栈实现前序遍历)
思路:用栈来存放二叉树的节点, 节点的左子树为空时删除该节点并将父节点的右子树添加进栈
func preorderTraversal(root: TreeNode?) -> [Int] {
var res = [Int]()
var stack = [TreeNode]()
var node = root
while !stack.isEmpty || node != nil {
if node != nil {
res.append(node!.val)
stack.append(node!)
node = node!.left
} else {
node = stack.removeLast().right
}
}
return res
}
用栈实现中序遍历二叉树
func cenOrderTraversal(root: TreeNode?) -> [Int] {
var res = [Int]()
var stack = [TreeNode]()
var node = root
while !stack.isEmpty || node != nil {
if node != nil {
stack.append(node!)
node = node!.left
} else {
node = stack.removeLast()
res.append(node!.val)
node = node?.right
}
}
return res
}
层级遍历二叉树
func levelOrder(root: TreeNode?) -> [[Int]] {
var res = [[Int]]()
var queue = [TreeNode]()
if let root = root {
queue.append(root)
}
while queue.count > 0 {
let size = queue.count
var level = [Int]()
for _ in 0 ..< size {
let node = queue.removeFirst()
level.append(node.val)
if let left = node.left {
queue.append(left)
}
if let right = node.right {
queue.append(right)
}
}
res.append(level)
}
return res
}
链接如下:
故胤道长_二叉树讲解
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