Preorder Traversal

Visit the root before each subtree, producing root-left-right order.

Algorithm

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

Basic Implementation

basic.c

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct Node { int value; struct Node* left; struct Node* right; } Node;
Node* node_new(int value, Node* left, Node* right) {
    Node* node = (Node*)malloc(sizeof(Node));
    node->value = value; node->left = left; node->right = right; return node;
}
void append(char* out, const char* text) { strcat(out, text); }
void render(Node* node, char* out) {
    char buf[16];
    if (node == NULL) { append(out, "_"); return; }
    sprintf(buf, "%d", node->value); append(out, buf);
    if (node->left != NULL || node->right != NULL) {
        append(out, "("); render(node->left, out); append(out, ","); render(node->right, out); append(out, ")");
    }
}
Node* sample_tree(void) {
    return node_new(4, node_new(2, node_new(1, NULL, NULL), node_new(3, NULL, NULL)),
                       node_new(6, node_new(5, NULL, NULL), node_new(7, NULL, NULL)));
}
void print_list(int* values, int n) {
    printf("[");
    for (int i = 0; i < n; i++) { if (i) printf(", "); printf("%d", values[i]); }
    printf("]\n");
}
void preorder(Node* node, int* output, int* n) { if (!node) return; output[(*n)++] = node->value; preorder(node->left, output, n); preorder(node->right, output, n); }
int main(void) { int output[7]; int n = 0; preorder(sample_tree(), output, &n); print_list(output, n); }

Complexity

Time: O(n)
Space: O(h) recursion stack

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.

preorder Preorder records the current node before visiting left and right subtrees.