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();
		}
	}

}