数据结构--Morris遍历--二叉树的遍历

二叉树——二叉树的遍历（递归与迭代）

Morris遍历

利用Morris遍历实现二叉树的先序， 中序， 后续遍历， 时间复杂度O(N)， 额外空间复杂度O(1)。

如果一个结点有左孩子，则回到该结点两次，否则只回到一次，而且当第二次回到该结点时，左子树已经遍历完了

Morris遍历规则：

1.来到的当前结点即为cur，如果cur没有左孩子，则cur向右移动，即cur = cur.right

2.如果cur有左孩子，则找到cur左子树上最右的结点，记为mostRight

　　①如果mostRight的右指针指向null（第一次遍历到cur时），则将其指向当前结点cur，cur向左移动  cur = cur.left

　　②如果mostRight的右指针指向cur（第二次回到cur时），则让其指向null，cur向右移动  cur = cur.right

package binaryTree.traverse;

/**
 * Created by Skye on 2018/5/3.
 *
 * 当前结点cur
 * 1.如果cur没有左孩子，则cur = cur.right
 * 2.如果cur有左孩子，则找到cur左孩子的最右结点mostRight
 *   1）如果mostRight.right == null 则让mostRight.right = cur，cur = cur.left
 *   2）如果mostRight.right == cur，则让mostRight.right = null，cur= cur.right
 */
public class MorrisTraversal {

    public static void morris(Tree node){
        if(node == null) return;
        Tree cur = node;
        while(cur != null){
            Tree mostRight = cur.left;
            if(mostRight != null){
                while(mostRight.right != null && mostRight != cur){
                    mostRight = mostRight.right;
                }
                if(mostRight.right == null){
                    mostRight.right = cur;
                    cur = cur.left;
                    continue;
                }else{
                    mostRight.right = null;
                }
            }
            cur = cur.right;
        }
    }

    public static void morrisIn(Tree node){
        if(node == null) return;
        Tree cur = node;
        while(cur != null){
            Tree 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.print(cur.val + " ");
            cur = cur.right;
        }
    }

    public static void morrisPre(Tree tree){
        if(tree == null) return;
        Tree cur = tree;
        while(cur != null){
            Tree 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.print(cur.val + " ");
                    cur = cur.left;
                    continue;
                }else{
                    mostRight.right = null;
                }
            } else{
                System.out.print(cur.val + " ");
            }

            cur = cur.right;
        }
    }

    public static void morrisPost(Tree node){
        if(node == null) return;
        Tree cur = node;
        while(cur != null){
            Tree 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;
                    printEdge(cur.left);
                }
            }
            cur = cur.right;
        }
        printEdge(node);
    }

    public static void printEdge(Tree cur) {
        if(cur == null) return;

        int val = cur.val;
        printEdge(cur.right);
        System.out.print(val + " ");
    }

}

　　

注：最后的反向输出一排右子树的方法：

1.采用链表逆向输出

public static void printEdge(Tree cur) {
        if(cur == null) return;

        int val = cur.val;
        printEdge(cur.right);
        System.out.print(val + " ");
    }

　　

2.先将链表反转，然后输出，最后在反转回来。

public static void printEdge1(Tree cur){
        Tree tail = reverse(cur);
        Tree node = tail;
        while(node != null){
            System.out.print(node.val + " ");
            node = node.right;
        }

        reverse(tail);
    }

    public static Tree reverse(Tree node){
        Tree pre = null;

        while(node != null){
            Tree next = node.right;
            node.right = pre;
            pre = node;
            node = next;
        }
        return pre;
    }