在二叉树中找到一个节点的后继节点
【题目】 现在有一种新的二叉树节点类型如下:
public class Node {
public int value;
public Node left;
public Node right;
public Node parent;
public Node(int data) {
this.value = data; }}
该结构比普通二叉树节点结构多了一个指向父节点的parent指针。假设有一棵Node类型的节点组成的二叉树,树中每个节点的parent指针都正确地指向 自己的父节点,头节点的parent指向null。只给一个在二叉树中的某个节点 node,请实现返回node的后继节点的函数。在二叉树的中序遍历的序列中, node的下一个节点叫作node的后继节点。
/*
* 后继节点是中序遍历的后面的节点
* 如果一个节点有右子树,后继节点就是右子树上最左的节点
* 如果X节点没有右子树,找X的父节点,直到找到x为某个节点的左子树
*/
public class 二叉树的后继节点 { 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 null; } if(node.right != null) { return getLeftMost(node.right); }else { Node pa = node.parent; while(pa != null && pa.left != node) { node = pa; pa = node.parent; } return pa; } } 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)); } }