二叉树的遍历
目录
- 递归遍历
- 非递归遍历
- Morris遍历
1. 递归遍历
递归版遍历只要当前节点不为null，就可以三次回到当前节点。
public static void preOrderRecur(Node head) {if (head == null) {return;}System.out.print(head.value + " ");preOrderRecur(head.left);preOrderRecur(head.right);}public static void inOrderRecur(Node head) {if (head == null) {return;}inOrderRecur(head.left);System.out.print(head.value + " ");inOrderRecur(head.right);}public static void posOrderRecur(Node head) {if (head == null) {return;}posOrderRecur(head.left);posOrderRecur(head.right);System.out.print(head.value + " ");}
2. 非递归遍历
其中后序遍历打印顺序为左右中，由先序遍历打印顺序为中左右，所以可以对先序遍历改进为中右左（改变添加顺序），添加到另外一个栈中，最后遍历打印就是左右中顺序了。
public static void preOrderUnRecur(Node head) {System.out.println("pre-order: ");while (head != null) {Stack<Node> stack = new Stack<>();stack.push(head);while (!stack.isEmpty()) {Node node = stack.pop();System.out.println(node.value + " ");if (node.right != null) {stack.push(node.right);}if (node.left != null) {stack.push(node.left);}}}}public static void inOrderUnRecur(Node head) {System.out.print("in-order: ");if (head != null) {Stack<Node> stack = new Stack<>();while (!stack.isEmpty() || head != null) {if (head != null) {stack.push(head.left);head = head.left;} else {head = stack.pop();System.out.println(head.value + " ");head = head.right;}}}System.out.println();}public static void posOrderUnRecur1(Node head) {System.out.print("pos-order: ");if (head != null) {Stack<Node> s1 = new Stack<>();Stack<Node> s2 = new Stack<>();s1.push(head);while (!s1.isEmpty()) {head = s1.pop();s2.push(head);if (head.left != null) {s1.push(head.left);}if (head.right != null) {s1.push(head.right);}while (!s2.isEmpty()) {System.out.print(s2.pop().value + " ");}}}System.out.println();}
3. Morris遍历
Morris遍历法，能以时间复杂度O(N)，空间复杂度O(1)实现二叉树的遍历。
程序流程：
假设指针cur指向当前节点，cur从头结点开始。
- 如果cur无左孩子，则cur = cur.right；
- 如果cur有左孩子，则找到cur左子树上最右的节点，记为mostRight，分为以下两种情况：
- 若mostRight的right指针为null，则让其指向cur，且cur = cur.left;
- 若mostRight的right指针指向cur，则让其指回null，且cur = cur.right。
假设二叉树如下：
12 34 5 6 7
则Morris遍历顺序为：1,2,4,2,5,1,3,6,3,7
Morris 遍历代码实现：
public static void morrisIn(Node head) {if (head == null) {return;}Node cur = head;Node mostRight = null;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;}}cur = cur.right;}}
特点：
- 当某个节点有左子树，则会到达该节点两次，如果没有左子树，只会到达一次。
- 当第二次回到某个节点时，它的左子树已遍历完。
实质：
Morris遍历是利用左子树最右节点的指针指向null或指向自己来标记是第一次来到该节点还是第二次来到该节点。
Morris遍历改先序遍历
在以下两种情况下打印节点:
- 当节点没有左子树时，打印当前节点；
- 当节点有左子树时并且第一次访问时打印该节点，就是当mosrRight的右指针为null时。
public static void morrisPre(Node head) {if (head == null) {return;}Node cur = head;Node mostRight = null;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.print(cur.value + " ");cur = cur.left;continue;} else {mostRight.right = null;}} else {System.out.print(cur.value + " ");}cur = cur.right;}System.out.println();}
Morris遍历改中序遍历
在以下两种情况下打印节点:
- 当节点没有左子树时，说明第一次和第二次是重合在一起的，打印当前节点即可。
- 当节点有左子树时，那么需要处理完左子树再打印，即当前节点准备右移时打印。
public static void morrisIn(Node head) {if (head == null) {return;}Node cur = head;Node mostRight = null;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.print(cur.value + " ");cur = cur.right;}System.out.println();}
Morris遍历改后序遍历
在以下两种情况下打印节点:
- 只有当到达某个节点两次时逆序打印该节点左子树的右边界;
- 在代码的最后逆序打印整棵树的右边界，而逆序的过程就和单向链表的反转过程类似。
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();}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;}