Construct Binary Tree from Inorder and Postorder

Root pick from postorder + inorder range partitioning (right first)

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Postorder Ptr-1
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Python Code

7            nonlocal post_ptr
8            if left > right:
9                return None
10 
●            root_val = postorder[post_ptr]
12            post_ptr -= 1
......
15 
16            root.right = build(mid + 1, right)
17            root.left = build(left, mid - 1)
18            return root
19
Current Line (11): Pick root from postorder

Tree Structure

3791520
Operation:Pick Root

Build Progress

Post Ptr

-1

Inorder Pivot

-

Traversal Inputs

Inorder: [9, 3, 15, 20, 7]

Postorder: [9, 15, 7, 20, 3]

Range: [0..-1]

Created Order

Created nodes appear here...
Step 1: Pick Root

Build Stack

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Step Explanation

Line 11: pick_root

Pick 3 from postorder and split inorder at index 1

  • > Meaning: Current postorder pointer gives the root value.
  • > Why: Postorder ends with Root, so scanning backward yields root first.
  • > Next: Create root and split inorder range.
  • > Progress snapshot: 0/26 steps, 0 nodes created.
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