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题目要求:
如标题(Node除了left,right,还有一个指向父节点的指针)
什么前驱和后继节点? 在二叉树的中序遍历中,一个节点的前一个节点,就是该节点的前驱,一个节点的下一个节点就是该节点的后继。
如何寻找一个节点的前驱?
如果该节点有左子树,那么该节点的前驱就是该节点左子树中最右边的节点;如果该节点没有左子树,从当前节点开始往上寻找,直到当前节点是其父节点的右孩子,那么这个父节点,就是起始节点的前驱。
如果寻找一个节点的后继?
如果该节点有右子树,则该节点的后继是该节点的右子树中最左边的元素;如果该节点没有右子树,从当前节点开始往上寻找,直到当前的节点是其父节点的左孩子,那么这个父节点就是起始节点的后继。
代码实现 :
package com.isea.brush.tree;
/**
* 寻找一个节点的后继节点和前驱节点
*/
public class SuccessorNode {
private static class Node {
private Integer value;
private Node left;
private Node right;
private Node parent;
public Node(int data) {
this.value = data;
}
}
/**
* 传入一个node,返回该node的后继节点
*
* @param node
* @return
*/
public static Node getSuccessorNode(Node node) {
if (node == null) {
return null;
}
if (node.right != null) {
return getMostLeft(node.right);
} else {
Node parent = node.parent;
while (parent != null && node != parent.left) {
node = parent;
parent = node.parent;
}
return parent;
}
}
/**
* 返回node为根节点的最左边节点
*
* @param node
* @return
*/
private static Node getMostLeft(Node node) {
while (node.left != null) {
node = node.left;
}
return node;
}
/**
* 返回node节点的前驱节点
*
* @param node
* @return
*/
public static Node getPioneerNode(Node node) {
if (node == null) {
return node;
}
if (node.left != null) {
return getMostRight(node.left);
} else {
Node parent = node.parent;
while (parent != null && node != parent.right) {
node = parent;
parent = node.parent;
}
return parent;
}
}
/**
* 返回以node为根节点的最右边节点
*
* @param node
* @return
*/
private static Node getMostRight(Node node) {
while (node.right != null) {
node = node.right;
}
return node;
}
public static void main(String[] args) {
Node head = new Node(6);
head.parent = null;
head.left = new Node(3);
head.left.parent = head;
head.left.left = new Node(1);
head.left.left.parent = head.left;
head.left.left.right = new Node(2);
head.left.left.right.parent = head.left.left;
head.left.right = new Node(4);
head.left.right.parent = head.left;
head.left.right.right = new Node(5);
head.left.right.right.parent = head.left.right;
head.right = new Node(9);
head.right.parent = head;
head.right.left = new Node(8);
head.right.left.parent = head.right;
head.right.left.left = new Node(7);
head.right.left.left.parent = head.right.left;
head.right.right = new Node(10);
head.right.right.parent = head.right;
Node test = head.left.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.left.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.left.right.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.left.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.left;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right;
System.out.println(test.value + " next: " + getSuccessorNode(test).value);
test = head.right.right; // 10's next is null
System.out.println(test.value + " next: " + getSuccessorNode(test));
}
}