【编程】二叉树问题

1。由先序和中序建立二叉树。
2。由中序和后序建立二叉树。
3。二叉树的非递归先序，中序和后序遍历

import java.util.Scanner;import java.util.Stack;class Node {    Node left;    Node right;    char value;    Node() {}    Node(char value) {        left = null;        right = null;        this.value = value;     }}public class BuidTree {    /**     * 由先序和中序建立二叉树     * @param A 先序遍历     * @param B 中序遍历     * @param L1 先序遍历下标起点     * @param H1 先序遍历下标终点      * @param L2 中序遍历下标起点     * @param R2 中序遍历下标终点     */    public static Node PreInCreate(char A[], char B[], int L1, int H1, int L2, int H2) {        Node root = new Node();        root.value = A[L1];        int i;        for(i = L2; B[i] != root.value; i++);        int llen = i - L2;        int rlen = H2 - i;        if(llen > 0) {            root.left = PreInCreate(A, B, L1+1, L1+llen, L2, L2+llen-1);        } else {            root.left = null;        }        if(rlen > 0) {            root.right = PreInCreate(A, B, H1-rlen+1, H1, H2-rlen+1, H2);        } else {            root.right = null;        }        return root;    }    /**     * 由中序和后序建立二叉树     * @param A 中序遍历     * @param B 后序遍历     * @param L1 中序遍历起点下标     * @param H1 中序遍历终点下标     * @param L2 后序遍历起点下标     * @param H2 后序遍历终点下标     * @return     */    public static Node InPostCreate(char A[], char B[], int L1, int H1, int L2, int H2) {        Node root = new Node(B[H2]);        int i;        for(i = L1; A[i] != root.value && i <= H1; i++);        int llen = i - L1;        int rlen = H1 - i;        if(llen > 0) {            root.left = InPostCreate(A, B, L1, L1+llen-1, L2, L2+llen-1);        } else {            root.left = null;        }        if(rlen > 0) {            root.right = InPostCreate(A, B, i+1, H1, H2-rlen, H2-1);        } else {            root.right = null;        }        return root;    }    //先序遍历    public static void preOrder(Node head) {        if(head == null) return;        System.out.print(head.value + "  ");        preOrder(head.left);        preOrder(head.right);    }    //非递归先序遍历     public static void preOrderUnRecurion(Node head) {        System.out.println("非递归先序遍历:");        if(head != null) {            Stack<Node> stack = new Stack<>();            stack.add(head);            while(!stack.isEmpty()) {                head = stack.pop();                System.out.print(head.value + " ");                if(head.right != null) {                    stack.add(head.right);                }                if(head.left != null)  {                    stack.add(head.left);                }            }        }    }    //中序遍历    public static void inOrder(Node head) {        if(head == null) return;        inOrder(head.left);        System.out.print(head.value + "  ");        inOrder(head.right);    }    public static void inOrderUnRecurion(Node head) {        System.out.println("非递归中序遍历:");        if(head != null) {            Stack<Node> stack = new Stack<>();            while(!stack.isEmpty() || head != null) {                if(head != null) {                    stack.add(head);                    head = head.left;                } else {                    head = stack.pop();                    System.out.print(head.value + " ");                    head = head.right;                }            }        }        System.out.println();    }    //后序遍历    public static void posOrder(Node head) {        if(head==null) {            return;        }        posOrder(head.left);        posOrder(head.right);        System.out.print(head.value + "  ");    }    public static void posOrderUnRecurion(Node head) {        System.out.println("非递归后序遍历: ");        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();    }    public static void testPreInCreate() {        String A, B;        Scanner scanner = new Scanner(System.in);        A = scanner.nextLine();        B = scanner.nextLine();        Node root = PreInCreate(A.toCharArray(), B.toCharArray(), 0, A.length()-1, 0, B.length()-1);        System.out.println("先序遍历:");        preOrder(root);        System.out.println();        System.out.println("中序遍历:");        inOrder(root);        System.out.println();        System.out.println("后序遍历:");        posOrder(root);    }    public static void testInPostCreate() {        String A, B;        Scanner scanner = new Scanner(System.in);        A = scanner.nextLine();        B = scanner.nextLine();        Node root = InPostCreate(A.toCharArray(), B.toCharArray(), 0, A.length()-1, 0, B.length()-1);        System.out.println(A + " " + B);        System.out.println("先序遍历:");        //preOrder(root);        preOrderUnRecurion(root);        System.out.println();        System.out.println("中序遍历:");        //inOrder(root);        inOrderUnRecurion(root);        System.out.println();        System.out.println("后序遍历:");        //posOrder(root);        posOrderUnRecurion(root);    }    public static void main(String[] args) {        //由先序和中序建立二叉树        //testPreInCreate();        System.out.println();        //由中序和后序建立二叉树        testInPostCreate();    }}