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June 28, 2017 04:29
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Inorder successor of binary search tree - need to compile.
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using System; | |
class Solution | |
{ | |
class Node | |
{ | |
public int value; | |
public Node left; | |
public Node right; | |
public Node parent; | |
} | |
// given 5, find 9 | |
// test case 2: given node 14 | |
// test case 3: give node 25 | |
// test case 4: given node 9 | |
// test case 5: given node 20 | |
// test case 6: given node 11 | |
// time complexity - BST O(logN) - O(N) - space - linked list O(N) | |
static Node findInOrderSuccessor(Node inputNode) | |
{ | |
// your code goes here | |
if(inputNode == null) | |
{ | |
return null; // ? | |
} | |
// assume that node at least has one child | |
bool hasRightChild = inputNode.right != null; // false, false | |
if(hasRightChild) | |
{ | |
return getMinimum(inputNode.right); // ? node 12 , return node 11 | |
} | |
else | |
{ | |
// find parent node which is bigger than givenNode | |
var givenValue = inputNode.value; // 5, 14, 25 | |
var iterate = inputNode; | |
while(iterate.parent != null && iterate.parent.value < givenValue) // 9 > 5, 12 < 14, 20 > 14 | |
{ | |
iterate = iterate.parent; // node 12, node 9 | |
} | |
return iterate.parent; | |
/* | |
// uttermost parent node, iterate node has no parent node | |
if(iterate.parent != null && iterate.parent.value > givenValue) | |
{ | |
return iterate.parent; // return node 9 , return node 20, return 12 | |
} | |
else | |
{ | |
return null; // return null, | |
} | |
*/ | |
} | |
} | |
private static Node getMinimum(Node root) | |
{ | |
var minimum = root; // node 12 | |
while(minimum.left != null) | |
{ | |
minimum = minimum.left; // node 11 | |
} | |
return minimum; // return node 11 | |
} | |
static void Main(string[] args) | |
{ | |
// test findInOrderSuccessor here | |
} | |
} |
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