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Let's assume that we have a binary search tree with node Y that hasn't a right child and for whom a successor exists in the tree.

I want to prove that if I insert a node X into the tree and node X is greater than node Y and less than the successor of node Y then X will be added as the right child of Y.

Intuitively, this makes sense to me but my question is how do we prove this formally?

user96199
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2 Answers2

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Due to the lack of definitions in the question, let us pick up a couple of definitions that are needed for a formal proof.

A binary rooted tree is a BST if its nodes are visited in its natural order by the inorder traversal. Inserting a new node into a BST means adding a new edge so that the new node becomes the child of some old node and the tree remains a BST.

Let us prove a useful lemma.
(Uniqueness of insertion into BST) Let $T$ be a BST. Let node $X$ be a new node that is not equal to any node in $T$. Then there exists a unique way to insert $X$ into $T$ so that $T$ remains a BST with the same root, where the insertion must add one edge to $T$ (without modifying the existing edges).

Proof by mathematical induction on $n$, the number of vertices in $T$.

  • The base case when $n=1$. If the new node is less than the root node, it has to inserted as the left child of the root node; otherwise, it has to be inserted as the right child of the root node.
  • Assume the lemma is true for BST whose number of vertices is smaller than $n$. Suppose $T$ is a BST of $n$ vertices with node $R$ as its root node. Suppose $X$ is a new node that is not equal to any node of $T$. There are two cases.
    • $X$ is less than $R$. If $R$ has no left child, then $X$ has to be inserted as the left child of $R$. Otherwise, $R$'s left subtree is a BST (because of the recursive nature of the inorder traversal). $T$ has to be inserted into $R$'s left subtree. By induction hypothesis, there exists a unique way to insert $X$ into $R$'s left subtree.
    • $X$ is greater than $R$. Just switch every "left" to "right" in the above case.

Proof is done. As you have indicated, this is intuitively so obvious!

Now let us prove a proposition that is slightly more general than what you want to prove.
(Right child insertion) Let $T$ be a BST with node $Y$ that does not have a right child. Let node $X$ be a new node larger than $Y$ such that any node that is larger than $Y$, if exists, must be larger than $X$. Then node $X$, if inserted, will be inserted as the right child of $Y$.

Proof. Let $T'$ be the binary tree obtained from $T$ when $X$ is added as the right child of $Y$. Compare the inorder traversal on $T$ and that on $T'$. By the definition of inorder traversal, the difference between them can happen only when each traverses the subtree rooted at $Y$.

Both traversals will visit the same left subtree of $Y$ first. Then both will visit $Y$. When it is time to visit the right subtree of $Y$, the former traversal has nothing to do since $Y$ has no right child in $T$ while the latter traversal will visit one more node, $X$ since $Y$ has $X$ as its right child in $T'$.

Since the nodes in $T$ is visited in its natural order by the former traversal, the nodes in $T'$ is visited in its natural order by the latter traversal as $X$ is the very next node that is larger than $Y$. By definition, $T'$ is a BST. According to the above uniqueness lemma, $X$ must be inserted as the right child of $Y$. Q.E.D.

John L.
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Maybe this is not the "proof" you're looking for, but here's an attempt to convince anyone that $X$ will be inserted in the binary-search tree (BST) as $Y$'s right child.

The "proof" relies on the binary-search tree property and on the definition and correctness of the "insertion" and "successor" functions of a BST.

You start with the definition of a successor of a node $y$ in a BST:

  1. If a node $y$ has a right subtree, then the successor of $y$, call it $z$, is the minimum node of that right subtree

  2. Otherwise, it is the first ancestor of $y$, lets call it $z$, such that $y$ falls in the left subtree of $z$.

  3. If $z$ is not found in the previous two points, then $z$ does not exist, which implies that $y$ is the greatest element in the BST.

However, we also need to look at the pseudocode of the insertion function.

function insertion(T, x):
    // T is the binary-search tree
    // x is the element which needs to inserted in T
    if is_empty(T):
        T.root = x
    else:
        c = T.root  // c is the current node.
        p = c.parent  // p is the parent of c

    while is_not_null(c):
        p = c
        if x < c:
            c = c.left
        else:
            c = c.right
    if x < p:
        p.left = x
    else:
        p.right = x
    x.parent = p

Since, in your case, we know that $T$ is not empty, then the else block will be always executed in any run of the insertion algorithm. In the while loop, we go down the tree, until we find a node that is null (and c represents this null node). However, note that we keep track of the parent of that null node c, i.e. p. We know that p is going to be the parent of x, but we still do not know if x needs to be the left or right child of p, given that x could either be respectively smaller or greater than p. Once that is determined, using the if-else block after the while loop, we can also set the parent of x to be p.

I will not prove that this insertion algorithm is correct, that is, it maintains the BST property. We assume it is correct. Here's the "proof" then.

We know that $Y$ does not have a right child, but it has a successor. Therefore, the successor must be the first ancestor of $Y$, call it $Z$, such that $Y$ falls in the left subtree of $Z$. We know that $Z$ exists (by assumption). If you insert $X$ in the binary-search tree $T$, such that $Y < X < Z$, then $X$ will be the next successor of $Y$ (by definition).

Given that $Y < X < Z$, at some point during the insertion algorithm, the condition x < c is false (because $X > Y$), where c is $Y$ and x is $X$. This must be true because there is no node between $Y$ and $X$ in the tree. Hence, in that case, $X$ will consequently be inserted in the tree as $Y$'s right child.

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