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Lachlan Jacob 5 jaren geleden
bovenliggende
commit
452714d230
3 gewijzigde bestanden met toevoegingen van 67 en 0 verwijderingen
  1. 39
    0
      700/main.c
  2. 23
    0
      700/problem.txt
  3. 5
    0
      700/run.sh

+ 39
- 0
700/main.c Bestand weergeven

@@ -0,0 +1,39 @@
#include <stdio.h>
#include <stdlib.h>

struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
};

struct TreeNode* searchBST(struct TreeNode*, int);
struct TreeNode* searchBST(struct TreeNode* root, int val){
if (root == NULL) {
return NULL;
} else if (root->val == val) {
return root;
} else if (root->val > val) {
return searchBST(root->left, val);
} else {
return searchBST(root->right, val);
}
}

int main() {
struct TreeNode *t2 = malloc(sizeof(struct TreeNode));
t2->val = 2;
t2->left = NULL;
t2->right = NULL;
struct TreeNode *t3 = malloc(sizeof(struct TreeNode));
t3->val = 3;
t3->left = NULL;
t3->right = NULL;
struct TreeNode *t = malloc(sizeof(struct TreeNode));
t->val = 1;
t->left = t2;
t->right = t3;
printf("Expected: 1 (as root of tree)\n");
printf("Got: %d\n", searchBST(t, 1)->val);
return 0;
}

+ 23
- 0
700/problem.txt Bestand weergeven

@@ -0,0 +1,23 @@
Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node's value equals the given value. Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.

For example,

Given the tree:
4
/ \
2 7
/ \
1 3

And the value to search: 2

You should return this subtree:

2
/ \
1 3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.


+ 5
- 0
700/run.sh Bestand weergeven

@@ -0,0 +1,5 @@
#!/bin/bash

gcc -o main main.c
./main
rm main

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