#include<iostream> bool is_in_array(int* begin, int* end, int target){ int mid = (end - begin)/2; if(*begin>target || *end<target) //value has to be between [begin,end] return false; // to continue if(begin[mid] == target) return true; if(begin[mid] > target) //look at the left subarray return is_in_array(begin,begin+mid-1,target); // (begin[mid] < target) look at the right subarray return is_in_array(begin+mid+1,end,target); } bool find_in_array(int arr[] ,int target,int size){ int* begin = arr; int* end = begin+size-1; bool result = is_in_array(begin,end,target); return result; } void test_is_in_array(bool expected, int arr[],int target,int size){ bool result = find_in_array(arr,target,size); std::cout<<(result==expected?"PASS: ":"FAIL: ") <<" result => "<<result<<" expected =>" <<expected<<std::endl; } int main( void ){ int arr1[] = {1,3,5,7,9,10}; int arr2[] = {1,3,7,9,10}; int arr3[] = {1,3,5,7,9,10,11,15,16,16,16,16,16,19,20}; int arr4[] = {1,3,5,7,9,10,15,20,25,50,1000,1000001}; int arr5[] = {1,3,5,7,9,10,11,12,13,14}; test_is_in_array(0,arr1,0,sizeof(arr1)/sizeof(*arr1)); test_is_in_array(1,arr2,7,sizeof(arr2)/sizeof(*arr2)); test_is_in_array(1,arr3,16,sizeof(arr3)/sizeof(*arr3)); test_is_in_array(1,arr4,50,sizeof(arr4)/sizeof(*arr4)); test_is_in_array(0,arr5,15,sizeof(arr5)/sizeof(*arr5)); test_is_in_array(1,arr5,14,sizeof(arr5)/sizeof(*arr5)); return 0; }
Source: LeetCode Solution: We do a preorder traversal add each node in the string followed by an arrow "->" if the current node is a leaf node (have no right or left child) then we add the string to the list, otherwise, we add an arrow and make a recursive call in a preorder fashion. The previous string is stored in stack and we can use it later when the call is returned to add another possible routes from root to the leaf nodes. Complexity Analysis Time complexity : O ( n ) . n nodes need to be traversed ( n represents the number of nodes in a given tree) . Solution: public List< String > binaryTreePaths( TreeNode root) { ArrayList< String > path = new ArrayList (); binaryPath(root, "" ,path); return path; } public static void binaryPath( TreeNode root, String s, List< String > path){ if (root == null ) return ; ...
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