Morris遍历

/**
 * Morris遍历：可以将非递归遍历中的空间复杂度降为O(1)。从而实现时间复杂度为O(N)，而空间复杂度为O(1)的精妙算法
 * <p>
 * 记作当前节点为cur。
 * 如果cur无左孩子，cur向右移动（cur=cur.right）
 * 如果cur有左孩子，找到cur左子树上最右的节点，记为mostright
 * 如果mostright的right指针指向空，让其指向cur，cur向左移动（cur=cur.left）
 * 如果mostright的right指针指向cur，让其指向空，cur向右移动（cur=cur.right）
 * <p>
 * 实现以上的原则，即实现了morris遍历。
 * <p>
 * 实质：建立一种机制，对于没有左子树的节点只到达一次，对于有左子树的节点会到达两次
 */
public class Morris {

    public static void morris(Node head) {
        Node cur = head;
        Node mostRight;
        while (cur != null) {
            System.out.println(cur.value);
            mostRight = cur.left;
            if (mostRight != null) {  // 左子树不为空
                while (mostRight.right != null && mostRight.right != cur) {
                    mostRight = mostRight.right;
                }
                if (mostRight.right == null) {  // 第一次访问该节点
                    mostRight.right = cur;
                    cur = cur.left;
                    continue;
                } else { // 第一次访问该节点
                    mostRight.right = null;
                }
            }
            cur = cur.right;
        }
    }

    public static void morrisIn(Node head) {
        Node cur = head;
        Node mostRight;
        while (cur != null) {
            mostRight = cur.left;
            if (mostRight != null) {
                while (mostRight.right != null && mostRight.right != cur) {
                    mostRight = mostRight.right;
                }
                if (mostRight.right == null) {
                    mostRight.right = cur;
                    cur = cur.left;
                    continue;
                } else {
                    mostRight.right = null;
                }
            }
            System.out.println(cur.value);
            cur = cur.right;
        }
    }

    public static void morrisPre(Node head) {
        Node cur = head;
        Node mostRight;
        while (cur != null) {
            mostRight = cur.left;
            if (mostRight != null) {
                while (mostRight.right != null && mostRight.right != cur) {
                    mostRight = mostRight.right;
                }
                if (mostRight.right == null) {
                    mostRight.right = cur;
                    System.out.println(cur.value);
                    cur = cur.left;
                    continue;
                } else {
                    mostRight.right = null;
                }
            } else {
                System.out.println(cur.value);
            }
            cur = cur.right;
        }
    }

    public static void main(String[] args) {
        Node node = new Node(1);
        node.left = new Node(2);
        node.right = new Node(3);
        node.left.left = new Node(4);
        node.left.right = new Node(5);
        node.right.left = new Node(6);
        node.right.right = new Node(7);
        morris(node);
        System.out.println("=================");
        morrisIn(node);
        System.out.println("=================");
        morrisPre(node);
    }

    /**
     * 二叉树结构
     */
    public static class Node {

        public int value;

        public Node left;

        public Node right;

        public Node(int value) {
            this.value = value;
        }

    }

}

/* 如有意见或建议，欢迎评论区留言；如发现代码有误，欢迎批评指正 */