Trees

BST Search

Search a binary search tree for one present and one absent value.

Algorithm

The canonical tree is 4(2(1,3),6(5,7)), so this Go DSA implementation can be compared directly with the rest of the DSA track.

Basic Implementation

basic.go 
package main

import (
    "fmt"
    "strings"
)

type Node struct { value int; left *Node; right *Node }
func render(node *Node) string {
    if node == nil { return "_" }
    if node.left == nil && node.right == nil { return fmt.Sprintf("%d", node.value) }
    return fmt.Sprintf("%d(%s,%s)", node.value, render(node.left), render(node.right))
}
func sampleTree() *Node {
    return &Node{4, &Node{2, &Node{1, nil, nil}, &Node{3, nil, nil}}, &Node{6, &Node{5, nil, nil}, &Node{7, nil, nil}}}
}
func listString(values []int) string {
    parts := []string{}
    for _, value := range values { parts = append(parts, fmt.Sprintf("%d", value)) }
    return "[" + strings.Join(parts, ", ") + "]"
}
func search(root *Node, target int) bool { node := root; for node != nil { if target == node.value { return true }; if target < node.value { node = node.left } else { node = node.right } }; return false }
func main() { root := sampleTree(); if search(root, 5) { fmt.Println("5 found") } else { fmt.Println("5 not found") }; if search(root, 8) { fmt.Println("8 found") } else { fmt.Println("8 not found") } }

Complexity

  • Time: O(h) per search
  • Space: O(1) iterative

Implementation notes

  • Render tree structure explicitly instead of printing node objects.
  • The replay highlights the node, traversal state, queue, path, or search cursor that changes at each step.
search path A comparison chooses one subtree at each step, so whole branches are skipped.