在二叉树中找到一个节点的后继节点

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。

 

该结构比普通二又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。

思路：找节点x的后继节点

 

 代码： 

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;
        }
    }

    //求某一个节点的后继节点
    public static Node getSuccessorNode(Node node) {
        if (node == null) {
            return node;
        }
        if (node.right != null) { //有右子树
            return getLeftMost(node.right);
        } else { //无右子树
            Node parent = node.parent; //找到其父节点
            //while循环找当前节点是其父节点的左子树
            while (parent != null && parent.left != node) {  //当前节点是其父节点的右子树
                node = parent; //node往上走到parent的位置
                parent = node.parent; //parent来到node父节点的位置
            }
            //当前节点是其父节点的左子树  或者  父节点为空（node是整棵树最右节点时）时退出while循环
            return parent; //退出循环说明找到了后继节点
        }
    }

    //返回一个节点子树中最左的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));
    }
}
//中序遍历结果：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
 */