Question

1902. Depth of BST Given Insertion Order

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Description

You are given a 0-indexed integer array order of length n, a permutation of integers from 1 to n representing the order of insertion into a binary search tree.

A binary search tree is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node's key.
• The right subtree of a node contains only nodes with keys greater than the node's key.
• Both the left and right subtrees must also be binary search trees.

The binary search tree is constructed as follows:

• order[0] will be the root of the binary search tree.
• All subsequent elements are inserted as the child of any existing node such that the binary search tree properties hold.

Return _the depth of the binary search tree_.

A binary tree's depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

 

Example 1:

Input: order = [2,1,4,3]
Output: 3
Explanation: The binary search tree has a depth of 3 with path 2->3->4.

Example 2:

Input: order = [2,1,3,4]
Output: 3
Explanation: The binary search tree has a depth of 3 with path 2->3->4.

Example 3:

Input: order = [1,2,3,4]
Output: 4
Explanation: The binary search tree has a depth of 4 with path 1->2->3->4.

 

Constraints:

• n == order.length
• 1 <= n <= 105
• order is a permutation of integers between 1 and n.

Solutions

Solution 1

from sortedcontainers import SortedDict


class Solution:
  def maxDepthBST(self, order: List[int]) -> int:
    sd = SortedDict({0: 0, inf: 0, order[0]: 1})
    ans = 1
    for v in order[1:]:
      lower = sd.bisect_left(v) - 1
      higher = lower + 1
      depth = 1 + max(sd.values()[lower], sd.values()[higher])
      ans = max(ans, depth)
      sd[v] = depth
    return ans

class Solution {
  public int maxDepthBST(int[] order) {
    TreeMap<Integer, Integer> tm = new TreeMap<>();
    tm.put(0, 0);
    tm.put(Integer.MAX_VALUE, 0);
    tm.put(order[0], 1);
    int ans = 1;
    for (int i = 1; i < order.length; ++i) {
      int v = order[i];
      Map.Entry<Integer, Integer> lower = tm.lowerEntry(v);
      Map.Entry<Integer, Integer> higher = tm.higherEntry(v);
      int depth = 1 + Math.max(lower.getValue(), higher.getValue());
      ans = Math.max(ans, depth);
      tm.put(v, depth);
    }
    return ans;
  }
}