最近，我的问题被标记为重复，比如这个，即使它们并不是重复的。因此，让我从以下内容开始，然后我会解释我的问题。
为什么这个问题不是重复的？
我并不是在问如何在给定中序遍历和前序遍历的情况下创建二叉树。我正在寻求证明，即中序遍历+前序遍历定义了唯一的二叉树。
现在，来看原始问题。我去参加了一次面试，面试官问了我这个问题。我被卡住了，无法继续下去。:|
问题：给定一棵二叉树的中序遍历和前序遍历。证明只有一棵二叉树可以用给定的数据构成。换句话说，证明两棵不同的二叉树不可能具有相同的中序和前序遍历。假设树中的所有元素都是唯一的（感谢@envy_intelligence指出这个假设）。
我尝试用例子说服面试官，但面试官要求数学/直观的证明。有人能帮我证明吗？
从前序遍历开始。如果为空，则完成了，否则它有一个第一个元素r0，即树的根。现在在中序遍历中搜索r0。左子树将全部位于该点之前，右子树将全部位于该点之后。因此，您可以在该点处将中序遍历分为左子树的中序遍历il和右子树的中序遍历ir。
如果il为空，则剩余的前序遍历属于右子树，并且您可以归纳继续进行。如果ir为空，同样的事情会发生在另一侧。如果两者都不为空，则在剩余的前序遍历中查找ir的第一个元素。这将把它分成左子树的前序遍历和右子树的前序遍历。归纳是直接的。
如果有人对正式证明感兴趣，我（最终）成功地在Idris中产生了一个。然而，我没有花时间尝试使它非常易读，因此实际上很难读懂其中的大部分内容。我建议您主要查看顶级类型（即引理、定理和定义），并尽量避免陷入证明（术语）中。
首先一些前提：
module PreIn
import Data.List
%default total
现在是第一个真正的想法：二叉树。
data Tree : Type -> Type where
Tip : Tree a
Node : (l : Tree a) -> (v : a) -> (r : Tree a) -> Tree a
%name Tree t, u
data InTree : a -> Tree a -> Type where
AtRoot : x `InTree` (Node l x r)
OnLeft : x `InTree` l -> x `InTree` (Node l v r)
OnRight : x `InTree` r -> x `InTree` (Node l v r)
然后还有一系列可怕的引理，其中有几个是由Eric Mertens（glguy）在他在我的问题中的答案中提出的。
size : Tree a -> Nat
size Tip = Z
size (Node l v r) = size l + (S Z + size r)
onLeftInjective : OnLeft p = OnLeft q -> p = q
onLeftInjective Refl = Refl
onRightInjective : OnRight p = OnRight q -> p = q
onRightInjective Refl = Refl
inorder : Tree a -> List a
inorder Tip = []
inorder (Node l v r) = inorder l ++ [v] ++ inorder r
instance Uninhabited (Here = There y) where
uninhabited Refl impossible
instance Uninhabited (x `InTree` Tip) where
uninhabited AtRoot impossible
elemAppend : {x : a} -> (ys,xs : List a) -> x `Elem` xs -> x `Elem` (ys ++ xs)
elemAppend [] xs xInxs = xInxs
elemAppend (y :: ys) xs xInxs = There (elemAppend ys xs xInxs)
appendElem : {x : a} -> (xs,ys : List a) -> x `Elem` xs -> x `Elem` (xs ++ ys)
appendElem (x :: zs) ys Here = Here
appendElem (y :: zs) ys (There pr) = There (appendElem zs ys pr)
tThenInorder : {x : a} -> (t : Tree a) -> x `InTree` t -> x `Elem` inorder t
tThenInorder (Node l x r) AtRoot = elemAppend _ _ Here
tThenInorder (Node l v r) (OnLeft pr) = appendElem _ _ (tThenInorder _ pr)
tThenInorder (Node l v r) (OnRight pr) = elemAppend _ _ (There (tThenInorder _ pr))
listSplit_lem : (x,z : a) -> (xs,ys:List a) -> Either (x `Elem` xs) (x `Elem` ys)
-> Either (x `Elem` (z :: xs)) (x `Elem` ys)
listSplit_lem x z xs ys (Left prf) = Left (There prf)
listSplit_lem x z xs ys (Right prf) = Right prf
listSplit : {x : a} -> (xs,ys : List a) -> x `Elem` (xs ++ ys) -> Either (x `Elem` xs) (x `Elem` ys)
listSplit [] ys xelem = Right xelem
listSplit (z :: xs) ys Here = Left Here
listSplit {x} (z :: xs) ys (There pr) = listSplit_lem x z xs ys (listSplit xs ys pr)
mutual
inorderThenT : {x : a} -> (t : Tree a) -> x `Elem` inorder t -> InTree x t
inorderThenT Tip xInL = absurd xInL
inorderThenT {x} (Node l v r) xInL = inorderThenT_lem x l v r xInL (listSplit (inorder l) (v :: inorder r) xInL)
inorderThenT_lem : (x : a) ->
(l : Tree a) -> (v : a) -> (r : Tree a) ->
x `Elem` inorder (Node l v r) ->
Either (x `Elem` inorder l) (x `Elem` (v :: inorder r)) ->
InTree x (Node l v r)
inorderThenT_lem x l v r xInL (Left locl) = OnLeft (inorderThenT l locl)
inorderThenT_lem x l x r xInL (Right Here) = AtRoot
inorderThenT_lem x l v r xInL (Right (There locr)) = OnRight (inorderThenT r locr)
unsplitRight : {x : a} -> (e : x `Elem` ys) -> listSplit xs ys (elemAppend xs ys e) = Right e
unsplitRight {xs = []} e = Refl
unsplitRight {xs = (x :: xs)} e = rewrite unsplitRight {xs} e in Refl
unsplitLeft : {x : a} -> (e : x `Elem` xs) -> listSplit xs ys (appendElem xs ys e) = Left e
unsplitLeft {xs = []} Here impossible
unsplitLeft {xs = (x :: xs)} Here = Refl
unsplitLeft {xs = (x :: xs)} {ys} (There pr) =
rewrite unsplitLeft {xs} {ys} pr in Refl
splitLeft_lem1 : (Left (There w) = listSplit_lem x y xs ys (listSplit xs ys z)) ->
(Left w = listSplit xs ys z)
splitLeft_lem1 {w} {xs} {ys} {z} prf with (listSplit xs ys z)
splitLeft_lem1 {w} Refl | (Left w) = Refl
splitLeft_lem1 {w} Refl | (Right s) impossible
splitLeft_lem2 : Left Here = listSplit_lem x x xs ys (listSplit xs ys z) -> Void
splitLeft_lem2 {x} {xs} {ys} {z} prf with (listSplit xs ys z)
splitLeft_lem2 {x = x} {xs = xs} {ys = ys} {z = z} Refl | (Left y) impossible
splitLeft_lem2 {x = x} {xs = xs} {ys = ys} {z = z} Refl | (Right y) impossible
splitLeft : {x : a} -> (xs,ys : List a) ->
(loc : x `Elem` (xs ++ ys)) ->
Left e = listSplit {x} xs ys loc ->
appendElem {x} xs ys e = loc
splitLeft {e} [] ys loc prf = absurd e
splitLeft (x :: xs) ys Here prf = rewrite leftInjective prf in Refl
splitLeft {e = Here} (x :: xs) ys (There z) prf = absurd (splitLeft_lem2 prf)
splitLeft {e = (There w)} (y :: xs) ys (There z) prf =
cong $ splitLeft xs ys z (splitLeft_lem1 prf)
splitMiddle_lem3 : Right Here = listSplit_lem y x xs (y :: ys) (listSplit xs (y :: ys) z) ->
Right Here = listSplit xs (y :: ys) z
splitMiddle_lem3 {y} {x} {xs} {ys} {z} prf with (listSplit xs (y :: ys) z)
splitMiddle_lem3 {y = y} {x = x} {xs = xs} {ys = ys} {z = z} Refl | (Left w) impossible
splitMiddle_lem3 {y = y} {x = x} {xs = xs} {ys = ys} {z = z} prf | (Right w) =
cong $ rightInjective prf -- This funny dance strips the Rights off and then puts them
-- back on so as to change type.
splitMiddle_lem2 : Right Here = listSplit xs (y :: ys) pl ->
elemAppend xs (y :: ys) Here = pl
splitMiddle_lem2 {xs} {y} {ys} {pl} prf with (listSplit xs (y :: ys) pl) proof prpr
splitMiddle_lem2 {xs = xs} {y = y} {ys = ys} {pl = pl} Refl | (Left loc) impossible
splitMiddle_lem2 {xs = []} {y = y} {ys = ys} {pl = pl} Refl | (Right Here) = rightInjective prpr
splitMiddle_lem2 {xs = (x :: xs)} {y = x} {ys = ys} {pl = Here} prf | (Right Here) = (\Refl impossible) prpr
splitMiddle_lem2 {xs = (x :: xs)} {y = y} {ys = ys} {pl = (There z)} prf | (Right Here) =
cong $ splitMiddle_lem2 {xs} {y} {ys} {pl = z} (splitMiddle_lem3 prpr)
splitMiddle_lem1 : Right Here = listSplit_lem y x xs (y :: ys) (listSplit xs (y :: ys) pl) ->
elemAppend xs (y :: ys) Here = pl
splitMiddle_lem1 {y} {x} {xs} {ys} {pl} prf with (listSplit xs (y :: ys) pl) proof prpr
splitMiddle_lem1 {y = y} {x = x} {xs = xs} {ys = ys} {pl = pl} Refl | (Left z) impossible
splitMiddle_lem1 {y = y} {x = x} {xs = xs} {ys = ys} {pl = pl} Refl | (Right Here) = splitMiddle_lem2 prpr
splitMiddle : Right Here = listSplit xs (y::ys) loc ->
elemAppend xs (y::ys) Here = loc
splitMiddle {xs = []} prf = rightInjective prf
splitMiddle {xs = (x :: xs)} {loc = Here} Refl impossible
splitMiddle {xs = (x :: xs)} {loc = (There y)} prf = cong $ splitMiddle_lem1 prf
splitRight_lem1 : Right (There pl) = listSplit (q :: xs) (y :: ys) (There z) ->
Right (There pl) = listSplit xs (y :: ys) z
splitRight_lem1 {xs} {ys} {y} {z} prf with (listSplit xs (y :: ys) z)
splitRight_lem1 {xs = xs} {ys = ys} {y = y} {z = z} Refl | (Left x) impossible
splitRight_lem1 {xs = xs} {ys = ys} {y = y} {z = z} prf | (Right x) =
cong $ rightInjective prf -- Type dance: take the Right off and put it back on.
splitRight : Right (There pl) = listSplit xs (y :: ys) loc ->
elemAppend xs (y :: ys) (There pl) = loc
splitRight {pl = pl} {xs = []} {y = y} {ys = ys} {loc = loc} prf = rightInjective prf
splitRight {pl = pl} {xs = (x :: xs)} {y = y} {ys = ys} {loc = Here} Refl impossible
splitRight {pl = pl} {xs = (x :: xs)} {y = y} {ys = ys} {loc = (There z)} prf =
let rec = splitRight {pl} {xs} {y} {ys} {loc = z} in cong $ rec (splitRight_lem1 prf)
这些可怕的引理导致了以下关于中序遍历的定理，它们共同展示了在树中查找特定元素的方法与在其中序遍历中查找该元素的方法之间的一一对应关系。
---------------------------
-- tThenInorder is a bijection from ways to find a particular element in a tree
-- and ways to find that element in its inorder traversal. `inorderToFro`
-- and `inorderFroTo` together demonstrate this by showing that `inorderThenT` is
-- its inverse.
||| `tThenInorder t` is a retraction of `inorderThenT t`
inorderFroTo : {x : a} -> (t : Tree a) -> (loc : x `Elem` inorder t) -> tThenInorder t (inorderThenT t loc) = loc
inorderFroTo Tip loc = absurd loc
inorderFroTo (Node l v r) loc with (listSplit (inorder l) (v :: inorder r) loc) proof prf
inorderFroTo (Node l v r) loc | (Left here) =
rewrite inorderFroTo l here in splitLeft _ _ loc prf
inorderFroTo (Node l v r) loc | (Right Here) = splitMiddle prf
inorderFroTo (Node l v r) loc | (Right (There x)) =
rewrite inorderFroTo r x in splitRight prf
||| `inorderThenT t` is a retraction of `tThenInorder t`
inorderToFro : {x : a} -> (t : Tree a) -> (loc : x `InTree` t) -> inorderThenT t (tThenInorder t loc) = loc
inorderToFro (Node l v r) (OnLeft xInL) =
rewrite unsplitLeft {ys = v :: inorder r} (tThenInorder l xInL)
in cong $ inorderToFro _ xInL
inorderToFro (Node l x r) AtRoot =
rewrite unsplitRight {x} {xs = inorder l} {ys = x :: inorder r} (tThenInorder (Node Tip x r) AtRoot)
in Refl
inorderToFro {x} (Node l v r) (OnRight xInR) =
rewrite unsplitRight {x} {xs = inorder l} {ys = v :: inorder r} (tThenInorder (Node Tip v r) (OnRight xInR))
in cong $ inorderToFro _ xInR
许多相同的引理可以用来证明前序遍历的相应定理:
preorder : Tree a -> List a
preorder Tip = []
preorder (Node l v r) = v :: (preorder l ++ preorder r)
tThenPreorder : (t : Tree a) -> x `InTree` t -> x `Elem` preorder t
tThenPreorder Tip AtRoot impossible
tThenPreorder (Node l x r) AtRoot = Here
tThenPreorder (Node l v r) (OnLeft loc) = appendElem _ _ (There (tThenPreorder _ loc))
tThenPreorder (Node l v r) (OnRight loc) = elemAppend (v :: preorder l) (preorder r) (tThenPreorder _ loc)
mutual
preorderThenT : (t : Tree a) -> x `Elem` preorder t -> x `InTree` t
preorderThenT {x = x} (Node l x r) Here = AtRoot
preorderThenT {x = x} (Node l v r) (There y) = preorderThenT_lem (listSplit _ _ y)
preorderThenT_lem : Either (x `Elem` preorder l) (x `Elem` preorder r) -> x `InTree` (Node l v r)
preorderThenT_lem {x = x} {l = l} {v = v} {r = r} (Left lloc) = OnLeft (preorderThenT l lloc)
preorderThenT_lem {x = x} {l = l} {v = v} {r = r} (Right rloc) = OnRight (preorderThenT r rloc)
splitty : Right pl = listSplit xs ys loc -> elemAppend xs ys pl = loc
splitty {pl = Here} {xs = xs} {ys = (x :: zs)} {loc = loc} prf = splitMiddle prf
splitty {pl = (There x)} {xs = xs} {ys = (y :: zs)} {loc = loc} prf = splitRight prf
preorderFroTo : {x : a} -> (t : Tree a) -> (loc : x `Elem` preorder t) ->
tThenPreorder t (preorderThenT t loc) = loc
preorderFroTo Tip Here impossible
preorderFroTo (Node l x r) Here = Refl
preorderFroTo (Node l v r) (There loc) with (listSplit (preorder l) (preorder r) loc) proof spl
preorderFroTo (Node l v r) (There loc) | (Left pl) =
rewrite sym (splitLeft {e=pl} (preorder l) (preorder r) loc spl)
in cong {f = There} $ cong {f = appendElem (preorder l) (preorder r)} (preorderFroTo _ _)
preorderFroTo (Node l v r) (There loc) | (Right pl) =
rewrite preorderFroTo r pl in cong {f = There} (splitty spl)
preorderToFro : {x : a} -> (t : Tree a) -> (loc : x `InTree` t) -> preorderThenT t (tThenPreorder t loc) = loc
preorderToFro (Node l x r) AtRoot = Refl
preorderToFro (Node l v r) (OnLeft loc) =
rewrite unsplitLeft {ys = preorder r} (tThenPreorder l loc)
in cong {f = OnLeft} (preorderToFro l loc)
preorderToFro (Node l v r) (OnRight loc) =
rewrite unsplitRight {xs = preorder l} (tThenPreorder r loc)
in cong {f = OnRight} (preorderToFro r loc)
loc1和loc2是在树t中找到值x的方式，那么loc1必须等于loc2，则树t是可逆的。InjTree : Tree a -> Type
InjTree t = (x : a) -> (loc1, loc2 : x `InTree` t) -> loc1 = loc2
InjList : List a -> Type
InjList xs = (x : a) -> (loc1, loc2 : x `Elem` xs) -> loc1 = loc2
||| If a tree is injective, so is its preorder traversal
treePreInj : (t : Tree a) -> InjTree t -> InjList (preorder t)
treePreInj {a} t it x loc1 loc2 =
let foo = preorderThenT {a} {x} t loc1
bar = preorderThenT {a} {x} t loc2
baz = it x foo bar
in rewrite sym $ preorderFroTo t loc1
in rewrite sym $ preorderFroTo t loc2
in cong baz
||| If a tree is injective, so is its inorder traversal
treeInInj : (t : Tree a) -> InjTree t -> InjList (inorder t)
treeInInj {a} t it x loc1 loc2 =
let foo = inorderThenT {a} {x} t loc1
bar = inorderThenT {a} {x} t loc2
baz = it x foo bar
in rewrite sym $ inorderFroTo t loc1
in rewrite sym $ inorderFroTo t loc2
in cong baz
||| If a tree's preorder traversal is injective, so is the tree.
injPreTree : (t : Tree a) -> InjList (preorder t) -> InjTree t
injPreTree {a} t il x loc1 loc2 =
let
foo = tThenPreorder {a} {x} t loc1
bar = tThenPreorder {a} {x} t loc2
baz = il x foo bar
in rewrite sym $ preorderToFro t loc1
in rewrite sym $ preorderToFro t loc2
in cong baz
||| If a tree's inorder traversal is injective, so is the tree.
injInTree : (t : Tree a) -> InjList (inorder t) -> InjTree t
injInTree {a} t il x loc1 loc2 =
let
foo = tThenInorder {a} {x} t loc1
bar = tThenInorder {a} {x} t loc2
baz = il x foo bar
in rewrite sym $ inorderToFro t loc1
in rewrite sym $ inorderToFro t loc2
in cong baz
headsSame : {x:a} -> {xs : List a} -> {y : a} -> {ys : List a} -> (x :: xs) = (y :: ys) -> x = y
headsSame Refl = Refl
tailsSame : {x:a} -> {xs : List a} -> {y : a} -> {ys : List a} -> (x :: xs) = (y :: ys) -> xs = ys
tailsSame Refl = Refl
appendLeftCancel : {xs,ys,ys' : List a} -> xs ++ ys = xs ++ ys' -> ys = ys'
appendLeftCancel {xs = []} prf = prf
appendLeftCancel {xs = (x :: xs)} prf = appendLeftCancel {xs} (tailsSame prf)
lengthDrop : (xs,ys : List a) -> drop (length xs) (xs ++ ys) = ys
lengthDrop [] ys = Refl
lengthDrop (x :: xs) ys = lengthDrop xs ys
lengthTake : (xs,ys : List a) -> take (length xs) (xs ++ ys) = xs
lengthTake [] ys = Refl
lengthTake (x :: xs) ys = cong $ lengthTake xs ys
appendRightCancel_lem : (xs,xs',ys : List a) -> xs ++ ys = xs' ++ ys -> length xs = length xs'
appendRightCancel_lem xs xs' ys eq =
let foo = lengthAppend xs ys
bar = replace {P = \b => length b = length xs + length ys} eq foo
baz = trans (sym bar) $ lengthAppend xs' ys
in plusRightCancel (length xs) (length xs') (length ys) baz
appendRightCancel : {xs,xs',ys : List a} -> xs ++ ys = xs' ++ ys -> xs = xs'
appendRightCancel {xs} {xs'} {ys} eq with (appendRightCancel_lem xs xs' ys eq)
| lenEq = rewrite sym $ lengthTake xs ys
in let foo : (take (length xs') (xs ++ ys) = xs') = rewrite eq in lengthTake xs' ys
in rewrite lenEq in foo
listPartsEqLeft : {xs, xs', ys, ys' : List a} ->
length xs = length xs' ->
xs ++ ys = xs' ++ ys' ->
xs = xs'
listPartsEqLeft {xs} {xs'} {ys} {ys'} leneq appeq =
rewrite sym $ lengthTake xs ys
in rewrite leneq
in rewrite appeq
in lengthTake xs' ys'
listPartsEqRight : {xs, xs', ys, ys' : List a} ->
length xs = length xs' ->
xs ++ ys = xs' ++ ys' ->
ys = ys'
listPartsEqRight leneq appeq with (listPartsEqLeft leneq appeq)
listPartsEqRight leneq appeq | Refl = appendLeftCancel appeq
thereInjective : There loc1 = There loc2 -> loc1 = loc2
thereInjective Refl = Refl
injTail : InjList (x :: xs) -> InjList xs
injTail {x} {xs} xxsInj v vloc1 vloc2 = thereInjective $
xxsInj v (There vloc1) (There vloc2)
splitInorder_lem2 : ((loc1 : Elem v (v :: xs ++ v :: ysr)) ->
(loc2 : Elem v (v :: xs ++ v :: ysr)) -> loc1 = loc2) ->
Void
splitInorder_lem2 {v} {xs} {ysr} f =
let
loc2 = elemAppend {x=v} xs (v :: ysr) Here
in (\Refl impossible) $ f Here (There loc2)
-- preorderLength and inorderLength could be proven using the bijections
-- between trees and their traversals, but it's much easier to just prove
-- them directly.
preorderLength : (t : Tree a) -> length (preorder t) = size t
preorderLength Tip = Refl
preorderLength (Node l v r) =
rewrite sym (plusSuccRightSucc (size l) (size r))
in cong {f=S} $
rewrite sym $ preorderLength l
in rewrite sym $ preorderLength r
in lengthAppend _ _
inorderLength : (t : Tree a) -> length (inorder t) = size t
inorderLength Tip = Refl
inorderLength (Node l v r) =
rewrite lengthAppend (inorder l) (v :: inorder r)
in rewrite inorderLength l
in rewrite inorderLength r in Refl
preInLength : (t : Tree a) -> length (preorder t) = length (inorder t)
preInLength t = trans (preorderLength t) (sym $ inorderLength t)
splitInorder_lem1 : (v : a) ->
(xsl, xsr, ysl, ysr : List a) ->
(xsInj : InjList (xsl ++ v :: xsr)) ->
(ysInj : InjList (ysl ++ v :: ysr)) ->
xsl ++ v :: xsr = ysl ++ v :: ysr ->
v `Elem` (xsl ++ v :: xsr) ->
v `Elem` (ysl ++ v :: ysr) ->
xsl = ysl
splitInorder_lem1 v [] xsr [] ysr xsInj ysInj eq locxs locys = Refl
splitInorder_lem1 v [] xsr (v :: ysl) ysr xsInj ysInj eq Here Here with (ysInj v Here (elemAppend (v :: ysl) (v :: ysr) Here))
splitInorder_lem1 v [] xsr (v :: ysl) ysr xsInj ysInj eq Here Here | Refl impossible
splitInorder_lem1 v [] xsr (y :: ysl) ysr xsInj ysInj eq Here (There loc) with (headsSame eq)
splitInorder_lem1 v [] xsr (v :: ysl) ysr xsInj ysInj eq Here (There loc) | Refl = absurd $ splitInorder_lem2 (ysInj v)
splitInorder_lem1 v [] xsr (x :: xs) ysr xsInj ysInj eq (There loc) locys with (headsSame eq)
splitInorder_lem1 v [] xsr (v :: xs) ysr xsInj ysInj eq (There loc) locys | Refl = absurd $ splitInorder_lem2 (ysInj v)
splitInorder_lem1 v (v :: xs) xsr ysl ysr xsInj ysInj eq Here locys = absurd $ splitInorder_lem2 (xsInj v)
splitInorder_lem1 v (x :: xs) xsr [] ysr xsInj ysInj eq (There y) locys with (headsSame eq)
splitInorder_lem1 v (v :: xs) xsr [] ysr xsInj ysInj eq (There y) locys | Refl = absurd $ splitInorder_lem2 (xsInj v)
splitInorder_lem1 v (x :: xs) xsr (z :: ys) ysr xsInj ysInj eq (There y) locys with (headsSame eq)
splitInorder_lem1 v (v :: xs) xsr (_ :: ys) ysr xsInj ysInj eq (There y) Here | Refl = absurd $ splitInorder_lem2 (ysInj v)
splitInorder_lem1 v (x :: xs) xsr (x :: ys) ysr xsInj ysInj eq (There y) (There z) | Refl = cong {f = ((::) x)} $
splitInorder_lem1 v xs xsr ys ysr (injTail xsInj) (injTail ysInj) (tailsSame eq) y z
splitInorder_lem3 : (v : a) ->
(xsl, xsr, ysl, ysr : List a) ->
(xsInj : InjList (xsl ++ v :: xsr)) ->
(ysInj : InjList (ysl ++ v :: ysr)) ->
xsl ++ v :: xsr = ysl ++ v :: ysr ->
v `Elem` (xsl ++ v :: xsr) ->
v `Elem` (ysl ++ v :: ysr) ->
xsr = ysr
splitInorder_lem3 v xsl xsr ysl ysr xsInj ysInj prf locxs locys with (splitInorder_lem1 v xsl xsr ysl ysr xsInj ysInj prf locxs locys)
splitInorder_lem3 v xsl xsr xsl ysr xsInj ysInj prf locxs locys | Refl =
tailsSame $ appendLeftCancel prf
一个简单的事实：如果一棵树是单射的，则其左子树和右子树也是单射的。
injLeft : {l : Tree a} -> {v : a} -> {r : Tree a} ->
InjTree (Node l v r) -> InjTree l
injLeft {l} {v} {r} injlvr x loc1 loc2 with (injlvr x (OnLeft loc1) (OnLeft loc2))
injLeft {l = l} {v = v} {r = r} injlvr x loc1 loc1 | Refl = Refl
injRight : {l : Tree a} -> {v : a} -> {r : Tree a} ->
InjTree (Node l v r) -> InjTree r
injRight {l} {v} {r} injlvr x loc1 loc2 with (injlvr x (OnRight loc1) (OnRight loc2))
injRight {l} {v} {r} injlvr x loc1 loc1 | Refl = Refl
如果t和u是二叉树，其中t是单射的，并且t和u具有相同的前序遍历和中序遍历，则t和u相等。
travsDet : (t, u : Tree a) -> InjTree t -> preorder t = preorder u -> inorder t = inorder u -> t = u
-- The base case--both trees are empty
travsDet Tip Tip x prf prf1 = Refl
-- Impossible cases: only one tree is empty
travsDet Tip (Node l v r) x Refl prf1 impossible
travsDet (Node l v r) Tip x Refl prf1 impossible
-- The interesting case. `headsSame presame` proves
-- that the roots of the trees are equal.
travsDet (Node l v r) (Node t y u) lvrInj presame insame with (headsSame presame)
travsDet (Node l v r) (Node t v u) lvrInj presame insame | Refl =
let
foo = elemAppend (inorder l) (v :: inorder r) Here
bar = elemAppend (inorder t) (v :: inorder u) Here
inlvrInj = treeInInj _ lvrInj
intvuInj : (InjList (inorder (Node t v u))) = rewrite sym insame in inlvrInj
inorderRightSame = splitInorder_lem3 v (inorder l) (inorder r) (inorder t) (inorder u) inlvrInj intvuInj insame foo bar
preInL : (length (preorder l) = length (inorder l)) = preInLength l
inorderLeftSame = splitInorder_lem1 v (inorder l) (inorder r) (inorder t) (inorder u) inlvrInj intvuInj insame foo bar
inPreT : (length (inorder t) = length (preorder t)) = sym $ preInLength t
preLenlt : (length (preorder l) = length (preorder t))
= trans preInL (trans (cong inorderLeftSame) inPreT)
presame' = tailsSame presame
baz : (preorder l = preorder t) = listPartsEqLeft preLenlt presame'
quux : (preorder r = preorder u) = listPartsEqRight preLenlt presame'
-- Putting together the lemmas, we see that the
-- left and right subtrees are equal
recleft = travsDet l t (injLeft lvrInj) baz inorderLeftSame
recright = travsDet r u (injRight lvrInj) quux inorderRightSame
in rewrite recleft in rewrite recright in Refl
如果树中的两个位置不相等，那么它们不包含相同的元素，那么就可以说树“没有重复项”。这可以使用NoDups类型来表示。
NoDups : Tree a -> Type
NoDups {a} t = (x, y : a) ->
(loc1 : x `InTree` t) ->
(loc2 : y `InTree` t) ->
Not (loc1 = loc2) ->
Not (x = y)
这个原因足以证明我们所需要的，因为有一个程序可以确定树中两条路径是否相等：
instance DecEq (x `InTree` t) where
decEq AtRoot AtRoot = Yes Refl
decEq AtRoot (OnLeft x) = No (\Refl impossible)
decEq AtRoot (OnRight x) = No (\Refl impossible)
decEq (OnLeft x) AtRoot = No (\Refl impossible)
decEq (OnLeft x) (OnLeft y) with (decEq x y)
decEq (OnLeft x) (OnLeft x) | (Yes Refl) = Yes Refl
decEq (OnLeft x) (OnLeft y) | (No contra) = No (contra . onLeftInjective)
decEq (OnLeft x) (OnRight y) = No (\Refl impossible)
decEq (OnRight x) AtRoot = No (\Refl impossible)
decEq (OnRight x) (OnLeft y) = No (\Refl impossible)
decEq (OnRight x) (OnRight y) with (decEq x y)
decEq (OnRight x) (OnRight x) | (Yes Refl) = Yes Refl
decEq (OnRight x) (OnRight y) | (No contra) = No (contra . onRightInjective)
这表明Nodups t意味着InjTree t：
noDupsInj : (t : Tree a) -> NoDups t -> InjTree t
noDupsInj t nd x loc1 loc2 with (decEq loc1 loc2)
noDupsInj t nd x loc1 loc2 | (Yes prf) = prf
noDupsInj t nd x loc1 loc2 | (No contra) = absurd $ nd x x loc1 loc2 contra Refl
最后，立即得出结论：NoDups t 完成了工作。
travsDet2 : (t, u : Tree a) -> NoDups t -> preorder t = preorder u -> inorder t = inorder u -> t = u
travsDet2 t u ndt = travsDet t u (noDupsInj t ndt)
splitLeft等相关内容。 - dfeuera,b,c,d,e,f,g。这告诉了您什么?a是树的根节点，这是先序遍历的定义所得出的。到目前为止，一切顺利。b,c向左走，d,e,f,g向右走等等。a的左子树中的元素将在中序遍历中在a之前出现，而任何在右子树中的元素将在a之后出现。同样，这是由中序遍历的定义推出的。c,b,a,d,e,f,g)。我们可以看到b和c在a之前出现，因此它们在左子树中，而d、e、f和g在右子树中。换句话说，在中序遍历中a的位置唯一地确定了哪些节点将在其左/右子树中。b,c/中序遍历c,b和先序遍历d,e,f,g/中序遍历d,e,f,g。我会问面试官一个关于重复元素的问题。如果两个“不同”的二叉树具有相同的前序遍历和中序遍历，则它们可能具有重复元素。
例如，考虑以下情况：
中序遍历：{12, 12} 前序遍历：{12, 12}
12 12
/ \
12 12