在二叉樹的中序遍歷的序列中，node的下一個節點叫作node的後繼節點。

該結構比普通二又樹節點結構多了一個指向父節點的parent指標。
假設有一棵Node型別的節點組成的二叉樹，樹中每個節點的parent指標都正確地指向自己的父節點，頭節點的parent指向null。
只給一個在二叉樹中的某個節點node，請實現返回node的後繼節點的函式。

思路：找節點x的後繼節點

程式碼：

package Algorithms.tree;

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;
        }
    }

    //求某一個節點的後繼節點
    public static Node getSuccessorNode(Node node) {
        if (node == null) {
            return node;
        }
        if (node.right != null
 
) { //有右子樹
            return getLeftMost(node.right);
        } else { //無右子樹
            Node parent = node.parent; //找到其父節點
            //while迴圈找當前節點是其父節點的左子樹
            while (parent != null && parent.left != node) {  //當前節點是其父節點的右子樹
                node = parent; //node往上走到parent的位置
                parent = node.parent; //
 
parent來到node父節點的位置
            }
            //當前節點是其父節點的左子樹  或者  父節點為空（node是整棵樹最右節點時）時退出while迴圈
            return parent; //退出迴圈說明找到了後繼節點
        }
    }

    //返回一個節點子樹中最左的Node
    public 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));
    }
}
//中序遍歷結果：1,2,3,4,5,6,7,8,9
/**
 * 1 next: 2
 * 2 next: 3
 * 3 next: 4
 * 4 next: 5
 * 5 next: 6
 * 6 next: 7
 * 7 next: 8
 * 8 next: 9
 * 9 next: 10
 * 10 next: null
 */