通过二叉树的先序序列和中序序列构建二叉树,然后再main函数中输入二叉树的先序序列为ABDGCEF,中序序列为DGBAECF,并通过后续遍历算法输出二叉树的后续序列GDBEFCA;
时间: 2023-10-29 07:00:16 浏览: 42
好的,这个问题可以使用递归的方法来解决。首先我们可以通过先序序列的第一个元素A来确定根节点,然后在中序序列中找到根节点的位置,这样就可以确定左子树和右子树的中序序列。接下来,我们可以根据左子树和右子树的中序序列长度,来确定先序序列中左子树和右子树的先序序列。然后递归地构建左子树和右子树,最终构建出整棵二叉树。
以下是C++代码实现:
```cpp
#include <iostream>
#include <string>
#include <unordered_map>
using namespace std;
struct TreeNode {
char val;
TreeNode *left;
TreeNode *right;
TreeNode(char x) : val(x), left(NULL), right(NULL) {}
};
TreeNode* buildTreeHelper(string preOrder, string inOrder, int preStart, int preEnd, int inStart, int inEnd, unordered_map<char, int>& map) {
if (preStart > preEnd || inStart > inEnd) {
return nullptr;
}
char rootVal = preOrder[preStart];
TreeNode* root = new TreeNode(rootVal);
int rootIndex = map[rootVal];
int leftSubTreeSize = rootIndex - inStart;
root->left = buildTreeHelper(preOrder, inOrder, preStart + 1, preStart + leftSubTreeSize, inStart, rootIndex - 1, map);
root->right = buildTreeHelper(preOrder, inOrder, preStart + leftSubTreeSize + 1, preEnd, rootIndex + 1, inEnd, map);
return root;
}
TreeNode* buildTree(string preOrder, string inOrder) {
if (preOrder.empty() || inOrder.empty()) {
return nullptr;
}
unordered_map<char, int> map;
for (int i = 0; i < inOrder.size(); i++) {
map[inOrder[i]] = i;
}
return buildTreeHelper(preOrder, inOrder, 0, preOrder.size() - 1, 0, inOrder.size() - 1, map);
}
void postOrder(TreeNode* root) {
if (root == nullptr) {
return;
}
postOrder(root->left);
postOrder(root->right);
cout << root->val;
}
int main() {
string preOrder = "ABDGCEF";
string inOrder = "DGBAECF";
TreeNode* root = buildTree(preOrder, inOrder);
postOrder(root);
return 0;
}
```
输出结果为:
```
GDBEFCA
```