二叉树前序:遍历顺序为,根节点、左子树、右子树;中序:遍历顺序为,左子树、根节点、右子树;后序:遍历顺序为,左子树、右子树、根节点
可以发现,二叉树前序中的第一个节点为树的根节点root,然后找出root在中序里面的位置,就可以把前序和中序分别划分为左、右子树两个部分,然后递归调用即可。
举个例子,
中序遍历HKDBEAIFCGJ
后序遍历KHDEBIFJGCA
首先,A肯定是二叉树的根节点,然后5在中序里面的位置是5号(从0开始),此位置前面的是左子树中的节点,右面的是右子树的节点,
即 KHDEB|| IFJGC || A , HKDBE || A || IFCGJ,对红色的左子树序列、蓝色的右子树序列继续上述过程,直至结束。
//二叉树 前序和中序得到后序
#include <iostream>
using namespace std;
typedef struct node
{
char key;
struct node *left;
struct node *right;
}treeNode;
char mid_order[100];
char post_order[100];
treeNode* construct_pre_order(int post_l, int post_r, int mid_l, int mid_r)
{
if (post_r - post_l < 0)
{
return NULL;
}
treeNode *root;
root = new treeNode;
root->key = post_order[post_r];
if (post_r == post_l)
{
root->left = NULL;
root->right = NULL;
return root;
}
int index;
for (index = mid_l; index <= mid_r; index++)
{
if (mid_order[index] == post_order[post_r])
break;
}
root->left = construct_pre_order(post_l, post_l+(index-mid_l)-1, mid_l, index-1);
root->right = construct_pre_order(post_l+(index-mid_l), post_r-1, index+1, post_r);
return root;
}
void pre_Order(treeNode *root)
{
if(root != NULL)
{
cout<<(root->key);
pre_Order(root->left);
pre_Order(root->right);
}
}
int main()
{
int n;
cout<<"输入序列的长度\n";
cin>>n;
cout<<"输入二叉树后序\n";
for (int i = 0; i < n; i++)
cin>>post_order[i];
cout<<"输入二叉树中序\n";
for (int i = 0; i < n; i++)
cin>>mid_order[i];
treeNode *root = construct_pre_order(0, n-1, 0, n-1);
cout<<"二叉树的前序为\n";
pre_Order(root);
printf("\n");
return 0;
}