首先约定:先序:根,左,右 中序:左,根,右 后序:左,右,根
递归实现
递归实现中,总是会通过一个节点三次,所以先序中序和后序的惟一区别就是打印的机会不同
public class Node {//这是Node的构造 public int value; public Node left; public Node right; public Node(int data) { this.value = data; }} public void preOdferRecur(Node head) { if (head == null) { return; } System.out.print(head.value + " ");//先序遍历,打印放在第一行 preOdferRecur(head.left); preOdferRecur(head.right); } public void inOdferRecur(Node head) { if (head == null) { return; } inOdferRecur(head.left); System.out.print(head.value + " ");//中序遍历,打印放在第二行 inOdferRecur(head.right); } public void posOdferRecur(Node head) { if (head == null) { return; } posOdferRecur(head.left); posOdferRecur(head.right); System.out.print(head.value + " ");//后序遍历,打印放在第三行 }
非递归实现
非递归先序遍历
思路:申请一个栈,压入头结点,而后弹出栈节点,打印进去,再将弹出节点的右孩子压入,左孩子压入,一直反复这个过程,直到栈为空
public void preOrderUnRecur(Node head){ if (head!=null){ Stack<Node> stack=new Stack<Node>(); stack.add(head); while(!stack.isEmpty()){ head=stack.pop(); System.out.printf(head.value+" "); if (head.right!=null){ stack.push(head.right); } if (head.left!=null){ stack.push(head.left); } } } }
非递归中序遍历
思路:中序是左,根,右,因而思考将左边界都压入栈,直到节点为空,则从栈中拿出一个打印,以后节点右移,若以后节点不为空,则压入栈,以后节点为左
- 申请一个栈,记为stack。初始时,令变量cur=head。
- 先把cur节点压入栈,对以cur节点为头结点的子树来说,顺次把左边界压入栈中,即不停地令cur=cur.left,而后反复步骤2
- 直到cur为空,此时从stack中弹出一个节点,记为node。打印node,并让cur=node.right,而后继续反复步骤2
- 当stack为空且cur为空,这个过程进行
`public void inOrderUnRecur(Node head){ 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.println(head.value+" "); head =head.right; } } } }
非递归后序遍历
思路:后序遍历是左,右,中,先序是中,左,右,将中左右变成中右左,在建设个栈将中右左压入,弹出即是后序遍历的秩序
public void posOrderUnRecur(Node head){ if (head!=null){ Stack<Node> stack1=new Stack<Node>(); Stack<Node> stack2=new Stack<Node>(); stack1.add(head); while(!stack1.isEmpty()){ head=stack1.pop(); stack2.push(head); if (head.left!=null){ stack1.push(head.left); } if (head.right!=null){ stack1.push(head.right); } } while(!stack2.isEmpty()){ System.out.printf(stack2.pop().value+" "); } } }
连贯地址:https://blog.csdn.net/weixin_...