InorderTraversal
约 347 字
题目
Github: InorderTraversal
实现二叉树中序遍历的类型版本。
const tree1 = {
val: 1,
left: null,
right: {
val: 2,
left: {
val: 3,
left: null,
right: null,
},
right: null,
},
} as const
type A = InorderTraversal<typeof tree1> // [1, 3, 2]解题思路
在二叉树的有序遍历中，我们遍历一个节点的子树，然后“访问”这个节点，然后遍历它的另 一个子树。通常，我们会先遍历左子树，然后再遍历节点的右子树。
A
/ \
B C
/ \
D E
In-order Traversal: D, B, E, A, C答案
interface TreeNode {
val: number
left: TreeNode | null
right: TreeNode | null
}
type InorderTraversal<
T extends TreeNode | null,
R extends TreeNode = NonNullable<T>
>
= T extends null
? []
: [...InorderTraversal<R['left']>, R['val'], ...InorderTraversal<R['right']>]验证
const tree1 = {
val: 1,
left: null,
right: {
val: 2,
left: {
val: 3,
left: null,
right: null,
},
right: null,
},
} as const
const tree2 = {
val: 1,
left: null,
right: null,
} as const
const tree3 = {
val: 1,
left: {
val: 2,
left: null,
right: null,
},
right: null,
} as const
const tree4 = {
val: 1,
left: null,
right: {
val: 2,
left: null,
right: null,
},
} as const
type cases = [
Expect<Equal<InorderTraversal<null>, []>>,
Expect<Equal<InorderTraversal<typeof tree1>, [1, 3, 2]>>,
Expect<Equal<InorderTraversal<typeof tree2>, [1]>>,
Expect<Equal<InorderTraversal<typeof tree3>, [2, 1]>>,
Expect<Equal<InorderTraversal<typeof tree4>, [1, 2]>>,
]