morris遍历

分手后的思念是犯贱 2022-05-09 14:33 113阅读 0赞

通常，实现二叉树的前序（preorder）、中序（inorder）、后序（postorder）遍历有两个常用的方法：一是递归(recursive)，二是使用栈实现的迭代版本(stack+iterative)。这两种方法都是O(n)的空间复杂度（递归本身占用stack空间或者用户自定义的stack）。

本文介绍空间O(1)的遍历方法。

上次文章讲到，我们经典递归遍历其实有三次访问当前节点的机会，就看你再哪次进行操作，而分成了三种遍历。

[https://blog.csdn.net/hebtu666/article/details/82853988][https_blog.csdn.net_hebtu666_article_details_82853988]

morris有两次访问节点的机会。

它省空间的原理是利用了大量叶子节点的没有用的空间，记录之前的节点，做到了返回之前节点这件事情。

我们不说先序中序后序，先说morris遍历的原则：

1、如果没有左孩子，继续遍历右子树

2、如果有左孩子，找到左子树最右节点。

1）如果最右节点的右指针为空（说明第一次遇到），把它指向当前节点，当前节点向左继续处理。

2）如果最右节点的右指针不为空（说明它指向之前结点），把右指针设为空，当前节点向右继续处理。

这就是morris遍历。

请手动模拟深度至少为3的树的morris遍历来熟悉流程。

先看代码：

定义结点：

public static class Node {
    		public int value;
    		Node left;
    		Node right;
    
    		public Node(int data) {
    			this.value = data;
    		}
    	}

先序：

（完全按规则写就好。）

//打印时机（第一次遇到）：发现左子树最右的孩子右指针指向空，或无左子树。
    	public static void morrisPre(Node head) {
    		if (head == null) {
    			return;
    		}
    		Node cur1 = head;
    		Node cur2 = null;
    		while (cur1 != null) {
    			cur2 = cur1.left;
    			if (cur2 != null) {
    				while (cur2.right != null && cur2.right != cur1) {
    					cur2 = cur2.right;
    				}
    				if (cur2.right == null) {
    					cur2.right = cur1;
    					System.out.print(cur1.value + " ");
    					cur1 = cur1.left;
    					continue;
    				} else {
    					cur2.right = null;
    				}
    			} else {
    				System.out.print(cur1.value + " ");
    			}
    			cur1 = cur1.right;
    		}
    		System.out.println();
    	}

morris在发表文章时只写出了中序遍历。而先序遍历只是打印时机不同而已，所以后人改进出了先序遍历。至于后序，是通过打印所有的右边界来实现的：对每个有边界逆序，打印，再逆序回去。注意要原地逆序，否则我们morris遍历的意义也就没有了。

完整代码：

public class MorrisTraversal {
    
    	
    	
    	public static void process(Node head) {
    		if(head == null) {
    			return;
    		}
    		
    		// 1
    		//System.out.println(head.value);
    		
    		
    		process(head.left);
    		
    		// 2
    		//System.out.println(head.value);
    		
    		
    		process(head.right);
    		
    		// 3
    		//System.out.println(head.value);
    	}
    	
    	
    	public static class Node {
    		public int value;
    		Node left;
    		Node right;
    
    		public Node(int data) {
    			this.value = data;
    		}
    	}
    //打印时机：向右走之前
    	public static void morrisIn(Node head) {
    		if (head == null) {
    			return;
    		}
    		Node cur1 = head;//当前节点
    		Node cur2 = null;//最右
    		while (cur1 != null) {
    			cur2 = cur1.left;
    			//左孩子不为空
    			if (cur2 != null) {
    				while (cur2.right != null && cur2.right != cur1) {
    					cur2 = cur2.right;
    				}//找到最右
    				//右指针为空，指向cur1，cur1向左继续
    				if (cur2.right == null) {
    					cur2.right = cur1;
    					cur1 = cur1.left;
    					continue;
    				} else {
    					cur2.right = null;
    				}//右指针不为空，设为空
    			}
    			System.out.print(cur1.value + " ");
    			cur1 = cur1.right;
    		}
    		System.out.println();
    	}
    //打印时机（第一次遇到）：发现左子树最右的孩子右指针指向空，或无左子树。
    	public static void morrisPre(Node head) {
    		if (head == null) {
    			return;
    		}
    		Node cur1 = head;
    		Node cur2 = null;
    		while (cur1 != null) {
    			cur2 = cur1.left;
    			if (cur2 != null) {
    				while (cur2.right != null && cur2.right != cur1) {
    					cur2 = cur2.right;
    				}
    				if (cur2.right == null) {
    					cur2.right = cur1;
    					System.out.print(cur1.value + " ");
    					cur1 = cur1.left;
    					continue;
    				} else {
    					cur2.right = null;
    				}
    			} else {
    				System.out.print(cur1.value + " ");
    			}
    			cur1 = cur1.right;
    		}
    		System.out.println();
    	}
    //逆序打印所有右边界
    	public static void morrisPos(Node head) {
    		if (head == null) {
    			return;
    		}
    		Node cur1 = head;
    		Node cur2 = null;
    		while (cur1 != null) {
    			cur2 = cur1.left;
    			if (cur2 != null) {
    				while (cur2.right != null && cur2.right != cur1) {
    					cur2 = cur2.right;
    				}
    				if (cur2.right == null) {
    					cur2.right = cur1;
    					cur1 = cur1.left;
    					continue;
    				} else {
    					cur2.right = null;
    					printEdge(cur1.left);
    				}
    			}
    			cur1 = cur1.right;
    		}
    		printEdge(head);
    		System.out.println();
    	}
    //逆序打印
    	public static void printEdge(Node head) {
    		Node tail = reverseEdge(head);
    		Node cur = tail;
    		while (cur != null) {
    			System.out.print(cur.value + " ");
    			cur = cur.right;
    		}
    		reverseEdge(tail);
    	}
    //逆序（类似链表逆序）
    	public static Node reverseEdge(Node from) {
    		Node pre = null;
    		Node next = null;
    		while (from != null) {
    			next = from.right;
    			from.right = pre;
    			pre = from;
    			from = next;
    		}
    		return pre;
    	}
    	public static void main(String[] args) {
    		Node head = new Node(4);
    		head.left = new Node(2);
    		head.right = new Node(6);
    		head.left.left = new Node(1);
    		head.left.right = new Node(3);
    		head.right.left = new Node(5);
    		head.right.right = new Node(7);
    
    		morrisIn(head);
    		morrisPre(head);
    		morrisPos(head);
    	}
    
    }

[https_blog.csdn.net_hebtu666_article_details_82853988]: https://blog.csdn.net/hebtu666/article/details/82853988