剑指 Offer 08. 二叉树的下一个节点
题目:
给定一颗二叉树和其中一个节点,如何找出中序遍历序列的下一个节点?树中的节点除了有两个分别指向左、右节点的指针,还有一个指向父节点的指针。
如果一个节点有右子树,那么它的下一个节点就是它的右子树中最左子节点。则,从右子节点出发一直沿着指向左子节点的指针,我们就可以找到它的下一个节点。
如果一个节点没有右子树
- 如果节点是它父节点的左子节点,那么它的下一个节点就是它的父节点
- 如果节点是它父节点的子节点,则我们沿着指向父节点的指针一直向上遍历,直到找到一个是它父节点的左子节点的节点。如果节点存在,那么这个节点的父节点就是我们要找的下一个节点。
/**
* Copyright (C), 2019-2021
* author candy_chen
* date 2021/1/19 20:22
* version 1.0
* Description: 在二叉树中找到一个节点的后继节点(后继节点是中序遍历的下一个节点)
*/
/**
* 如果一个二叉树有右子树,则后继节点就是右子树上最左的节点
* 若没有右子树,。。。。。。。
*/
public class SuccessorNode {
public static class Node{
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data){
this.value = data;
}
}
public static Node getSuccessorNode(Node node){
if (node == null){
return node;
}
if (node.right != null){//该节点有右子树
return getLeftMost(node.right);
}else{
Node parent = node.parent;
while (parent != null && parent.left != node){
node = parent;
parent = node.parent;
}
return parent;
}
}
/**
* 找到该树的最左边的节点
* @param node
* @return
*/
private static Node getLeftMost(Node node) {
if (node == null){
return node;
}
while (node.left != null){
node = node.left;
}
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));
}
}