标签:sso his bsp == str 遍历 node ati 先序遍历
用递归的方法实现前序遍历,中序遍历,后序遍历:
public static class Node
{ public int value; public Node left; public Node right; public Node(int data)
{ this.value = data; } } 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 + " "); }
用非递归的方法实现前序遍历,中序遍历,后序遍历:
public static void preOrderUnRecur(Node head) { System.out.print("pre-order: "); if (head != null) { Stack<Node> stack = new Stack<Node>(); stack.add(head); while (!stack.isEmpty()) { head = stack.pop(); System.out.print(head.value + " "); if (head.right != null) { stack.push(head.right); } if (head.left != null) { stack.push(head.left); } } } System.out.println(); } public static void inOrderUnRecur(Node head) { System.out.print("in-order: "); if (head != null) { Stack<Node> stack = new Stack<Node>(); while (!stack.isEmpty() || head != null) { if (head != null) { stack.push(head); head = head.left; } else { head = stack.pop(); System.out.print(head.value + " "); head = head.right; } } } System.out.println(); } public static void posOrderUnRecur1(Node head) //此方法和先序遍历类似,并使用了一个辅助栈
{ System.out.print("pos-order: "); if (head != null) { Stack<Node> s1 = new Stack<Node>(); Stack<Node> s2 = new Stack<Node>(); 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(); } public static void posOrderUnRecur2(Node h) { System.out.print("pos-order: "); if (h != null) { Stack<Node> stack = new Stack<Node>(); stack.push(h); Node c = null; while (!stack.isEmpty()) { c = stack.peek(); if (c.left != null && h != c.left && h != c.right) { stack.push(c.left); } else if (c.right != null && h != c.right) { stack.push(c.right); } else { System.out.print(stack.pop().value + " "); h = c; } } } System.out.println(); }
为什么用栈来实现遍历二叉树,而不用队列?
因为树是一个自上而下的结构,只有从上到下的路径,所以需要想一个能让它回去的路径的方法,那就是使用栈。
2、中序遍历后继节点:
如果一个节点X如果有右子树,那么X的后继节点一定是右子树的最左节点。如果X没有右子树,那么就看哪个节点的左子树是以X结尾的(一直往上找,找到某个节点是父亲节点的左孩子就停,那个父节点就是X节点的后继)。
public class SuccessorNode //获得后继节点 { public static class Node { public int value; public Node left; public Node right; public Node parent; public Node(int data) { this.value = data; } } public static Node getSuccessorNode(Node node) { if (node == null) { return node; } if (node.right != null) { return getLeftMost(node.right); } else { Node parent = node.parent; while (parent != null && parent.left != node) { node = parent; parent = node.parent; } return parent; } } public static Node getLeftMost(Node node) { if (node == null) { return node; } while (node.left != null) { node = node.left; } return node; }
3、树的序列化与反序列化(做成字符串,存文本)
序列化:
public static class Node { public int value; public Node left; public Node right; public Node(int data) { this.value = data; } } public static String serialByPre(Node head) { if (head == null) { return "#!"; } String res = head.value + "!"; res += serialByPre(head.left); res += serialByPre(head.right); return res; }
标签:sso his bsp == str 遍历 node ati 先序遍历
原文地址:https://www.cnblogs.com/roscangjie/p/11039638.html