下圖右邊就是左邊的樹的morris序

只要一個節點有左樹,該節點一定會到達兩次.

后序

public class Code_MorrisTraversal {public static class Node{public int value;Node left;Node right;public Node(int data){this.value = data;}}/** 4* 2 6* 1 3 5 7**/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);morrisPre(head);morrisIn(head);morrisPos(head);}public static void process(Node node){if(null == node){return;}// 1process(node.left);// 2process(node.right);// 3}public static void morris(Node head){if(null == head){return;}Node cur = head;Node mostRight = null;while(cur != null){mostRight = cur.left;if(null != mostRight){// 如果有左子樹while (null != mostRight.right && mostRight.right != cur){mostRight = mostRight.right;}// cur左樹上最右節點if(null == mostRight.right){mostRight.right = cur;cur = cur.left;continue;}else{mostRight.right = null;}}cur = cur.right;}}// 先序遍歷public static void morrisPre(Node head){if(null == head){return;}Node cur = head;Node mostRight = null;while(cur != null){mostRight = cur.left;if(null != mostRight){// 如果有左子樹while (null != mostRight.right && mostRight.right != cur){mostRight = mostRight.right;}// cur左樹上最右節點if(null == mostRight.right){System.out.println(cur.value); // 第一次碰到該節點，打印該節點mostRight.right = cur;cur = cur.left;continue;}else{mostRight.right = null;}}else{System.out.println(cur.value);}cur = cur.right;}}// 中序遍歷public static void morrisIn(Node head){if(null == head){return;}Node cur = head;Node mostRight = null;while(cur != null){mostRight = cur.left;if(null != mostRight){// 如果有左子樹while (null != mostRight.right && mostRight.right != cur){mostRight = mostRight.right;}// cur左樹上最右節點if(null == mostRight.right){mostRight.right = cur;cur = cur.left;continue;}else{mostRight.right = null;}}System.out.println(cur.value);cur = cur.right;}}// 后序遍歷public static void morrisPos(Node head){if(null == head){return;}Node cur = head;Node mostRight = null;while(cur != null){mostRight = cur.left;if(null != mostRight){// 如果有左子樹while (null != mostRight.right && mostRight.right != cur){mostRight = mostRight.right;}// cur左樹上最右節點if(null == mostRight.right){mostRight.right = cur;cur = cur.left;continue;}else{mostRight.right = null;printEdge(cur.left); // 該節點的左子數的右節點逆序}}cur = cur.right;}printEdge(head); // 頭節點的右節點逆序}public static void printEdge(Node node){Node tail = reverseEdge(node);Node cur = tail;while(null != cur){System.out.println(cur.value+" ");cur = cur.right;}reverseEdge(tail);}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;} }

總結

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