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Challenges Given the preorder and the inorder of a tree, output the postorder

Definitions A binary tree is either a null (leaf), or an object (node). A node contains a value (non-negative integer) and two pointers (left and right) to two separate binary trees. A binary tre...

2 answers · posted 4y ago by Hakerh400‭ · last activity 3y ago by radarek‭

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#2: Post edited by

Hakerh400‭ · 2020-12-02T03:10:19Z (about 4 years ago)

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### Definitions
A binary tree is either a `null` (leaf), or an object (node). A node contains a value (non-negative integer) and two pointers (left and right) to two separate binary trees.
A binary tree can be traversed in many different ways. Here we define **preorder**, **inorder** and **postorder**. We define them recursively using a pseudocode:
preorder(tree):
if tree is not a leaf:
visit(tree.value)
preorder(tree.left)
preorder(tree.right)
inorder(tree):
if tree is not a leaf:
inorder(tree.left)
visit(tree.value)
inorder(tree.right)
postorder(tree):
if tree is not a leaf:
postorder(tree.left)
postorder(tree.right)
visit(tree.value)
Each algorithm sequentially visits all nodes of the given tree.
### Task
Given the sequence of nodes that is obtained by applying the preorder algorithm to a binary tree and given the sequence of nodes that is obtained by applying the inorder algorithm to the same tree, output the sequence of nodes that would be obtained by applying the postorder algorithm to the same tree.
### Input
Two arrays of integers. The first one corresponds to the preorder and the second one corresponds to the inorder. Both arrays contain the same number of elements. Each element in an array is unique (there are no two distinct nodes with the same value).
### Output
A single array of integers corresponding to the postorder.
### Test cases
Input: [[], []]
Output: []
Input: [[0], [0]]
Output: [0]
Input: [[0, 1], [1, 0]]
Output: [1, 0]
Input: [[0, 1], [0, 1]]
Output: [1, 0]
Input: [[0, 1, 2],
[1, 0, 2]]
Output: [1, 2, 0]
Input: [[0, 1, 2, 3, 4],
[1, 2, 0, 4, 3]]
Output: [2, 1, 4, 3, 0]
Input: [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
[4, 3, 2, 1, 7, 6, 8, 5, 9, 10, 0, 12, 14, 13, 15, 11, 18, 17, 16, 19, 20]]
Output: [4, 3, 2, 7, 8, 6, 10, 9, 5, 1, 14, 15, 13, 12, 18, 17, 20, 19, 16, 11, 0]
### Scoring
The shortest program in each language wins.

### Definitions
A binary tree is either a `null` (leaf), or an object (node). A node contains a value (non-negative integer) and two pointers (left and right) to two separate binary trees.
A binary tree can be traversed in many different ways. Here we define **preorder**, **inorder** and **postorder**. We define them recursively using a pseudocode:
preorder(tree):
if tree is not a leaf:
visit(tree.value)
preorder(tree.left)
preorder(tree.right)
inorder(tree):
if tree is not a leaf:
inorder(tree.left)
visit(tree.value)
inorder(tree.right)
postorder(tree):
if tree is not a leaf:
postorder(tree.left)
postorder(tree.right)
visit(tree.value)
Each algorithm sequentially visits all nodes of the given tree.
### Task
Given the sequence of nodes that is obtained by applying the preorder algorithm to a binary tree and given the sequence of nodes that is obtained by applying the inorder algorithm to the same tree, output the sequence of nodes that would be obtained by applying the postorder algorithm to the same tree.
### Input
Two arrays of integers. The first one corresponds to the preorder and the second one corresponds to the inorder. Both arrays contain the same number of elements. Each element in an array is unique (there are no two distinct nodes with the same value).
### Output
A single array of integers corresponding to the postorder.
### Test cases
Input: [[], []]
Output: []
Input: [[0], [0]]
Output: [0]
Input: [[0, 1], [1, 0]]
Output: [1, 0]
Input: [[0, 1], [0, 1]]
Output: [1, 0]
Input: [[0, 1, 2],
[1, 0, 2]]
Output: [1, 2, 0]
Input: [[0, 1, 2, 3, 4],
[1, 2, 0, 4, 3]]
Output: [2, 1, 4, 3, 0]
Input: [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
[4, 3, 2, 1, 7, 6, 8, 5, 9, 10, 0, 12, 14, 13, 15, 11, 18, 17, 16, 19, 20]]
Output: [4, 3, 2, 7, 8, 6, 10, 9, 5, 1, 14, 15, 13, 12, 18, 17, 20, 19, 16, 11, 0]
### Reference implementation
[Try it online!](https://tio.run/##lZRLc5swEMfv@RTbk6FRGMtxnh43p/aYHnJkOGhsYdMQICDSZDL@7Ok@BDY06aQXENr//vahFb/Mk2lWdVa5k6Jc27dVWTQOTNPY2sESavvYZrUNJrIzCRdekaIxqGqrICvKEJbf4PUIQGy3aCvaPF8c9Vu5aQhnOiEg2bV1ASY2UW6LjdvCCehkgbbdgV9tVxQoVRBFkak3TbgniOLB3Ns7Z1b3P2rzYL14pOy0aVsQzr1UtkyphOUSJrTpsrKYwA1uXUMaT5M4jXWScB57740t0FkY6JjCjTcDb0amqvKXgArvErh@T0B8L@j4vhcxRlDcOe5D14kufkNVYgbxsOS@3i7f39sst4EOR7X79tBJBIwKFwOBRH8yeUJVknpUf22bNqdTRGFU2GcXoHgEeV2XhWVKa3c8P@TTg7I0IEGfGu9wMt0QUGd12Jv73jCx54A0I6rKKugzgMPa8Pww/NgLk3RZQVvyvetMHtc222DUXUbwkfEqavJsRd0NfdzdaGI3fkZopL7ub4hUjNXix2Gt07Cr8FZ4ctHIsZN9YdlgFDAVDEOaZpulruuBGLNibZ/RTJF4/TM9OCnPFxGhT/SQndvUeW@pdaoEOQhSZ5vtUCbEY@hx3VT7FmMKyi9fMpuvceAUcH8oYvKBjeN4419/h2rQa7NvsuHe0k3q2xsn76eF/xUfssJ7RNdevW@Z7S2G5mGckufiHytAPb2kBt@eUOYAm7NbvMkhRGtrqztXZyv3/bE1eZAGMc5ZnIT8WBx9LKN5xEcoz38qFf5WUaYViF4Wn3Hh9@ddFMwSxVpaeb/ZZ10VnCqYC2DGjDluMWbGijkr/gem4EzBuYILBZcKrtBEZrRrIqJCo0SjRqNIo0qjTKNuxu2VgBL8gkGXTPSgqefMPevMsy8967xnhWrPklzOBXHFPH2QyKmHeshsKhCG@VN4@wM)
### Scoring
The shortest program in each language wins.

#1: Initial revision by

Hakerh400‭ · 2020-11-30T13:53:30Z (about 4 years ago)

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Given the preorder and the inorder of a tree, output the postorder

### Definitions

A binary tree is either a `null` (leaf), or an object (node). A node contains a value (non-negative integer) and two pointers (left and right) to two separate binary trees.

A binary tree can be traversed in many different ways. Here we define **preorder**, **inorder** and **postorder**. We define them recursively using a pseudocode:

    preorder(tree):
      if tree is not a leaf:
        visit(tree.value)
        preorder(tree.left)
        preorder(tree.right)
    
    inorder(tree):
      if tree is not a leaf:
        inorder(tree.left)
        visit(tree.value)
        inorder(tree.right)
    
    postorder(tree):
      if tree is not a leaf:
        postorder(tree.left)
        postorder(tree.right)
        visit(tree.value)

Each algorithm sequentially visits all nodes of the given tree.

### Task

Given the sequence of nodes that is obtained by applying the preorder algorithm to a binary tree and given the sequence of nodes that is obtained by applying the inorder algorithm to the same tree, output the sequence of nodes that would be obtained by applying the postorder algorithm to the same tree.

### Input

Two arrays of integers. The first one corresponds to the preorder and the second one corresponds to the inorder. Both arrays contain the same number of elements. Each element in an array is unique (there are no two distinct nodes with the same value).

### Output

A single array of integers corresponding to the postorder.

### Test cases

    Input:  [[], []]
    Output: []
    
    Input:  [[0], [0]]
    Output: [0]
    
    Input:  [[0, 1], [1, 0]]
    Output: [1, 0]
    
    Input:  [[0, 1], [0, 1]]
    Output: [1, 0]
    
    Input:  [[0, 1, 2],
             [1, 0, 2]]
    Output: [1, 2, 0]
    
    Input:  [[0, 1, 2, 3, 4],
             [1, 2, 0, 4, 3]]
    Output: [2, 1, 4, 3, 0]
    
    Input:  [[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20],
             [4, 3, 2, 1, 7, 6, 8, 5, 9, 10, 0, 12, 14, 13, 15, 11, 18, 17, 16, 19, 20]]
    Output: [4, 3, 2, 7, 8, 6, 10, 9, 5, 1, 14, 15, 13, 12, 18, 17, 20, 19, 16, 11, 0]

### Scoring

The shortest program in each language wins.

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