已知二叉树前序，中序遍历，求后序遍历，java实现

  简单介绍一下思想，先看前序，前序遍历的第一个节点，就是该树的根。在中序中找到该根的位置，设为index，在中序遍历集合中，位于index之前的属于根的左子树，位于index之后的属于根的右子树。然后，对左右子数，遍历的调用该过程，就能将树结构还原。之后后序遍历也就不是问题了。

public class M {
public static class Tree{
public int value;
public Tree left;
public Tree right;
public Tree(int value) {
this.value=value;
}
}
public static void main(String[] args){
int[]preOrder={1,2,4,5,3,6,7};
int[]midOrder={4,2,5,1,6,3,7};
getBehindOrder(preOrder, midOrder);
}
public static void getBehindOrder(int[]pre,int[]mid){
if (pre==null||mid==null||pre.length!=mid.length) {
return;
}
Tree root=buildTree(pre, mid);
behindOrder(root);
}

public static void behindOrder(Tree root){//后序遍历二叉树
if (root==null) {
return;
}
if (root.left!=null) {
behindOrder(root.left);
}
if (root.right!=null) {
behindOrder(root.right);
}
System.out.print(root.value+" ");

}
public static Tree buildTree(int[]preOrder,int[]midOrder){//根据前序和中序，建立二叉树
int value=preOrder[0];
int length=preOrder.length;
Tree root=new Tree(value);
root.left=root.right=null;
if (preOrder.length==1) {
return root;
}
int index=0;
while(midOrder[index]!=value)
index++;//此处还要考虑index==length-1的情况
if (index>0) {
//中序中，根节点左边的节点都属于左子树
int[]leftSubPreOrder=new int[index];
for(int i=0;i<leftSubPreOrder.length;i++){
leftSubPreOrder[i]=preOrder[i+1];
}
int[]leftSubMidOrder=new int[index];
for(int i=0;i<leftSubMidOrder.length;i++){
leftSubMidOrder[i]=midOrder[i];
}
root.left=buildTree(leftSubPreOrder, leftSubMidOrder);
}

if(length-index-1>0){
int[]rightSubMidOrder=new int[length-index-1];
for(int i=0;i<rightSubMidOrder.length;i++){
rightSubMidOrder[i]=midOrder[i+index+1];
}
int[]rightSubPreOrder=new int[length-index-1];

for(int i=0;i<rightSubPreOrder.length;i++){
rightSubPreOrder[i]=preOrder[i+index+1];
}

root.right=buildTree(rightSubPreOrder, rightSubMidOrder);
}
return root;
}
}