給定一棵二叉樹的頭節點 head，完成二叉樹的先序、中序和后序遍歷。
如果二叉樹的節點數為 N，要求時間復雜度為 O(N)，額外空間復雜度為 O(1)

主要思想：莫爾斯遍歷，使用空閑右指針，回到后續節點，只需要關心一個節點的左子樹的最右節點是懸空還是指向后續節點

public static class Node{public int value;Node left;Node right;public Node(int data){this.value = data;} }public static void morrisIn(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;}}System.out.print(cur1.value + " ");cur1 = cur1.right;}System.out.println(); }public static void morrisPre(Node head){if(head == null){return;}Node cur1 = head;Node cur2 = null;while(cur != 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(); }public static void morrisPos(Node head){if(head == null){return;}Node cur1 = head;Node cur2 = null;while(cur1 != null){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(); }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); }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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