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

1. 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
2. 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));
    }

}