题目
题目链接
求和树转换算法
解题思路
-
基本步骤：
- 根据前序和中序遍历重建原始二叉树
- 将原始二叉树转换为求和树
- 输出求和树的中序遍历
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关键点：
- 满二叉树的特点：每个非叶子节点都有两个子节点
- 求和树节点值 = 左子树所有节点值之和 + 右子树所有节点值之和
- 叶子节点的求和树值为0
代码实现
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
using namespace std;// 树节点定义
struct TreeNode {int val;TreeNode* left;TreeNode* right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};// 构建二叉树
TreeNode* buildTree(vector<int>& preorder, int preStart, int preEnd,vector<int>& inorder, int inStart, int inEnd) {if (preStart > preEnd) return NULL;TreeNode* root = new TreeNode(preorder[preStart]);// 在中序遍历中找根节点位置int rootIndex = inStart;for (int i = inStart; i <= inEnd; i++) {if (inorder[i] == preorder[preStart]) {rootIndex = i;break;}}int leftSubtreeSize = rootIndex - inStart;// 递归构建左右子树root->left = buildTree(preorder, preStart + 1, preStart + leftSubtreeSize,inorder, inStart, rootIndex - 1);root->right = buildTree(preorder, preStart + leftSubtreeSize + 1, preEnd,inorder, rootIndex + 1, inEnd);return root;
}// 转换为求和树
int convertToSumTree(TreeNode* root) {if (root == NULL) return 0;int oldValue = root->val;// 递归计算左右子树的和int leftSum = convertToSumTree(root->left);int rightSum = convertToSumTree(root->right);// 更新当前节点的值为左右子树的和root->val = leftSum + rightSum;// 返回包含当前节点的子树所有节点值的和return oldValue + leftSum + rightSum;
}// 中序遍历
void inorderTraversal(TreeNode* root, vector<int>& result) {if (root == NULL) return;inorderTraversal(root->left, result);result.push_back(root->val);inorderTraversal(root->right, result);
}int main() {string line;// 读取前序遍历getline(cin, line);istringstream iss1(line);vector<int> preorder;int num;while (iss1 >> num) {preorder.push_back(num);}// 读取中序遍历getline(cin, line);istringstream iss2(line);vector<int> inorder;while (iss2 >> num) {inorder.push_back(num);}// 构建二叉树TreeNode* root = buildTree(preorder, 0, preorder.size() - 1,inorder, 0, inorder.size() - 1);// 转换为求和树convertToSumTree(root);// 获取求和树的中序遍历vector<int> result;inorderTraversal(root, result);// 输出结果for (int i = 0; i < result.size(); i++) {cout << result[i];if (i < result.size() - 1) cout << " ";}cout << endl;return 0;
}
import java.util.*;public class Main {static class TreeNode {int val;TreeNode left;TreeNode right;TreeNode(int val) {this.val = val;}}public static void main(String[] args) {Scanner sc = new Scanner(System.in);// 读取前序遍历String[] preStr = sc.nextLine().split(" ");int[] preorder = new int[preStr.length];for(int i = 0; i < preStr.length; i++) {preorder[i] = Integer.parseInt(preStr[i]);}// 读取中序遍历String[] inStr = sc.nextLine().split(" ");int[] inorder = new int[inStr.length];for(int i = 0; i < inStr.length; i++) {inorder[i] = Integer.parseInt(inStr[i]);}// 重建二叉树TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);// 转换为求和树convertToSumTree(root);// 输出求和树的中序遍历List<Integer> result = new ArrayList<>();inorderTraversal(root, result);// 打印结果for(int i = 0; i < result.size(); i++) {System.out.print(result.get(i));if(i < result.size() - 1) {System.out.print(" ");}}}// 构建二叉树private static TreeNode buildTree(int[] preorder, int preStart, int preEnd,int[] inorder, int inStart, int inEnd) {if(preStart > preEnd) return null;TreeNode root = new TreeNode(preorder[preStart]);// 在中序遍历中找到根节点位置int rootIndex = 0;for(int i = inStart; i <= inEnd; i++) {if(inorder[i] == preorder[preStart]) {rootIndex = i;break;}}int leftSubtreeSize = rootIndex - inStart;// 递归构建左右子树root.left = buildTree(preorder, preStart + 1, preStart + leftSubtreeSize,inorder, inStart, rootIndex - 1);root.right = buildTree(preorder, preStart + leftSubtreeSize + 1, preEnd,inorder, rootIndex + 1, inEnd);return root;}// 转换为求和树private static int convertToSumTree(TreeNode root) {if(root == null) return 0;// 获取原始值int oldValue = root.val;// 递归计算左右子树的和int leftSum = convertToSumTree(root.left);int rightSum = convertToSumTree(root.right);// 更新当前节点的值为左右子树的和root.val = leftSum + rightSum;// 返回包含当前节点的子树所有节点值的和return oldValue + leftSum + rightSum;}// 中序遍历private static void inorderTraversal(TreeNode root, List<Integer> result) {if(root == null) return;inorderTraversal(root.left, result);result.add(root.val);inorderTraversal(root.right, result);}
}
class TreeNode:def __init__(self, val=0):self.val = valself.left = Noneself.right = Nonedef buildTree(preorder, preStart, preEnd, inorder, inStart, inEnd):if preStart > preEnd:return None# 创建根节点root = TreeNode(preorder[preStart])# 在中序遍历中找到根节点位置rootIndex = inStartfor i in range(inStart, inEnd + 1):if inorder[i] == preorder[preStart]:rootIndex = ibreakleftSubtreeSize = rootIndex - inStart# 递归构建左右子树root.left = buildTree(preorder, preStart + 1, preStart + leftSubtreeSize,inorder, inStart, rootIndex - 1)root.right = buildTree(preorder, preStart + leftSubtreeSize + 1, preEnd,inorder, rootIndex + 1, inEnd)return rootdef convertToSumTree(root):if not root:return 0# 保存原始值oldValue = root.val# 递归计算左右子树的和leftSum = convertToSumTree(root.left)rightSum = convertToSumTree(root.right)# 更新当前节点的值为左右子树的和root.val = leftSum + rightSum# 返回包含当前节点的子树所有节点值的和return oldValue + leftSum + rightSumdef inorderTraversal(root, result):if not root:returninorderTraversal(root.left, result)result.append(root.val)inorderTraversal(root.right, result)def solve():# 读取输入preorder = list(map(int, input().split()))inorder = list(map(int, input().split()))# 构建二叉树root = buildTree(preorder, 0, len(preorder) - 1,inorder, 0, len(inorder) - 1)# 转换为求和树convertToSumTree(root)# 获取求和树的中序遍历result = []inorderTraversal(root, result)# 输出结果print(' '.join(map(str, result)))if __name__ == "__main__":solve()
算法分析
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时间复杂度：
- 构建二叉树：\(\mathcal{O}(n^2)\)
- 转换为求和树：\(\mathcal{O}(n)\)
- 总体：\(\mathcal{O}(n^2)\)
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空间复杂度：
- 递归调用栈：\(\mathcal{O}(h)\)，\(h\)为树高
- 存储结果：\(\mathcal{O}(n)\)
- 总体：\(\mathcal{O}(n)\)
