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find the node in k distance in a given tree list, solution 1 : shifting -> complexity = o (n)
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public class HelloWorld { | |
public static void main(String[] args) { | |
Node r = new Node(0); | |
r.left = new Node(1); | |
r.right = new Node(2); | |
r.left.left = new Node(3); | |
r.left.right = new Node(4); | |
r.left.left.left = new Node(5); | |
r.right.right = new Node(6); | |
r.right.left = new Node(7); | |
r.right.left.right = new Node(8); | |
find(r, 2, null, -1); | |
} | |
public static class Node { | |
Node parent; | |
Node left; | |
Node right; | |
int value; | |
public Node(int value) { | |
this.value = value; | |
} | |
} | |
public static void find(Node n, int k, int[] list, int last) { | |
if (list == null) { | |
list = new int[k]; | |
for (int i = 0; i < k; i++) { | |
list[i] = -1; | |
} | |
} | |
if (n.left == null && n.right == null) { | |
System.out.println("\nleaf : " + n.value); | |
if (list[k - 1] != -1) { | |
int i = last + 1 == k ? 0 : last + 1; | |
System.out.println("node in k distance : " + list[i]); | |
} | |
} | |
if (n.left != null) { | |
n.left.parent = n; | |
int[] temp = new int[k]; | |
temp = list.clone(); | |
int templast = getLastIndex(temp, last, k, n.value); | |
find(n.left, k, temp, templast); | |
} | |
if (n.right != null) { | |
n.right.parent = n; | |
int[] temp = new int[k]; | |
temp = list.clone(); | |
int templast = getLastIndex(temp, last, k, n.value); | |
find(n.right, k, temp, templast); | |
} | |
} | |
private static int getLastIndex(int[] temp, int last, int k, int value) { | |
int templast = last; | |
if (templast + 1 < k) { | |
temp[templast + 1] = value; | |
templast++; | |
} else { | |
temp[templast + 1 - k] = value; | |
templast = templast + 1 - k; | |
} | |
return templast; | |
} | |
} |
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