给定某二叉树三序遍历中的两个，我们即可以通过生成该二叉树，并遍历的方法，求出剩下的一序，具体代码如下package Tree;

import java.io.BufferedInputStream;

import java.util.*;

public class BT {

class Node{

Node l;//左儿子

Node r;//右儿子

char c;//结点字符

public Node(char c) {

this.c = c;

this.l = null;

this.r = null;

}

}

Node root;

char[] str1,str2;

public BT() {

root = null;

}

public void postOrder(Node n) {

if(n.l!=null) {

postOrder(n.l);

}

if(n.r!=null) {

postOrder(n.r);

}

System.out.print(n.c);

}

public void firstOrder() {

this.firstOrder(this.root);

}

public void firstOrder(Node n) {

System.out.print(n.c);

if(n.l!=null) {

firstOrder(n.l);

}

if(n.r!=null) {

firstOrder(n.r);

}

}

public void postOrder() {

this.postOrder(this.root);

}

public Node build(int s1,int e1,int s2,int e2) {

char c = str1[s1];

Node Root = new Node(c);

int index = 0;

for(int i = s2;i<=e2;i++) {

if(str2[i]==c) {

index = i;

break;

}

}

if(index!=s2) {//如果左子树不为空

Root.l = build(s1+1,s1+index-s2,s2,index-1);

}

if(index!=e2) {//如果右子树不为空

Root.r = build(s1+index-s2+1,e1,index+1,e2);

}

return Root;

}

public Node build1(int s1,int e1,int s2,int e2) {//中后序还原树

char c = str2[e2];

Node Root = new Node(c);

int index = 0;

for(int i = s1;i<=e1;i++) {

if(str1[i]==c) {

index = i;

break;

}

}

if(index!=s1) {

Root.l = build1(s1,index-1,s2,s2+index-s1-1);

}

if(index!=e1) {

Root.r = build1(index+1,e1,s2+index-s1+1,e2-1);

}

return Root;

}

public static void main(String[] args) {

String s1,s2 = null;

Scanner sc = new Scanner(new BufferedInputStream(System.in));

BT bt = new BT();

while(sc.hasNext()) {

s1 = sc.next();

s2 = sc.next();

bt.str1 = new char[s1.length()];

bt.str2 = new char[s1.length()];

bt.str1 = s1.toCharArray();

bt.str2 = s2.toCharArray();

bt.root = bt.build1(0, s1.length()-1, 0, s1.length()-1);

bt.firstOrder();

}

}

}其中build是已知前中序，生成二叉树；build1是已知中后序，生成二叉树.