在二叉树中找到一个节点的后继节点

名字是乱打的

发布于 2022-05-13 09:49:11

3650

发布于 2022-05-13 09:49:11

文章被收录于专栏：软件工程

【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。假设有一棵该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。node的上一个节点叫作node的钱去节点.,如某树遍历结果是5 1 4 3 8 7 9,那么1的后继结点就是4,1的前驱结点是5

第一种方法 : 很简单,中序遍历整个树,把结果存起来,查一下要找的数后面的值即可.但是这种时间复杂度比较高,每次需要遍历整个树

第二种方法 :其实一个结点的后继结点有这样一个规律

如果当前结点有右子树,则其后继结点是右子树的最左结点
如果当前结点没有右子树,则从父结点开始向上找,一直到当前结点是其父结点的左孩子时候停,那么当前结点的父结点就是其后继结点.

代码

package com.algorithm.practice.tree;

public class FindPosNodeFromTree {
    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 leftest(node.right);
        }else {
            Node parent=node.parent;
            while (parent!=null&&node!=parent.left){
                node=node.parent;
                parent=node.parent;
            }
            return parent;
        }

    }

    private static Node leftest(Node curr) {
        while (curr.left!=null){
            curr=curr.left;
        }
        return curr;
    }


    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));
    }

}

同理 如果找其前驱结点