Medium

Len Chen
1 min readSep 30, 2018

Problem

Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.

Note that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.

For example,

Given the tree:
4
/ \
2 7
/ \
1 3
And the value to insert: 5

You can return this binary search tree:

         4
/ \
2 7
/ \ /
1 3 5

This tree is also valid:

         5
/ \
2 7
/ \
1 3
\
4

Solution — Recursion

Just like #700, we can find the correct node on which should new node should be attached.

Complexity

By #700, finding correct position takes O(h) (or O(logn)) time if h denotes to the length of the longest path from root to leaves and n denotes to counts of nodes in this tree. And the same with space complexity, O(h) (or O(logn)).

Solution — Iteration

Just like #700.

Complexity

By #700,

  • time complexity: O(h) (or O(logn))
  • space complexity: O(1)

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Len Chen
Len Chen

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