Licking Exercise - 59 From Binary Search Trees to Greater Sum Trees

59 从二叉搜索树到更大和树

1.问题描述
给出二叉 搜索 树的根节点,该二叉树的节点值各不相同,修改二叉树,使每个节点 node 的新值等于原树中大于或等于 node.val The sum of the values ​​of all nodes of .

提醒一下,二叉搜索树满足下列约束条件：

节点的左子树仅包含键 小于 节点键的节点.

节点的右子树仅包含键 大于 节点键的节点.

左右子树也必须是二叉搜索树.

示例：

tree.png

输入：[4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]

输出：[30,36,21,36,35,26,15,33,8]

可使用以下main函数：

#include

#include

#include

#include

#include

#include

using namespace std;

struct TreeNode

{

int val;

TreeNode *left;

TreeNode *right;

TreeNode() : val(0), left(NULL), right(NULL) {}

TreeNode(int x) : val(x), left(NULL), right(NULL) {}

TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

};

TreeNode* inputTree()

{

int n,count=0;

char item[100];

cin>>n;

if (n==0)

    return NULL;

cin>>item;

TreeNode* root = new TreeNode(atoi(item));

count++;

queue<TreeNode*> nodeQueue;

nodeQueue.push(root);

while (count<n)

{

    TreeNode* node = nodeQueue.front();

    nodeQueue.pop();

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int leftNumber = atoi(item);

        node->left = new TreeNode(leftNumber);

        nodeQueue.push(node->left);

    }

    if (count==n)

        break;

    cin>>item;

    count++;

    if (strcmp(item,"null")!=0)

    {

        int rightNumber = atoi(item);

        node->right = new TreeNode(rightNumber);

        nodeQueue.push(node->right);

    }

}

return root;

}

void printTree(TreeNode* root) {

if (root == NULL) {

  return;

}

bool isFirst=true;

queue<TreeNode*> q;

q.push(root);

while(!q.empty()) {

    TreeNode* node = q.front();

    q.pop();

    if (node == NULL) {

      continue;

    }

    if (!isFirst)

        cout<<",";

    cout<<node->val;

    isFirst=false;

    q.push(node->left);

    q.push(node->right);

}

}

int main()

{

TreeNode* root;

root=inputTree();

TreeNode* res=Solution().bstToGst(root);

printTree(res);

}

2.输入说明
First enter the number of nodesn（注意,The nodes here include the ones in the titlenull空结点）

然后输入n个结点的数据,A node that needs to be filled with empty nodes,输入null.
3.输出说明
输出节点的信息,以逗号分隔
4.范例
输入
15
4 1 6 0 2 5 7 null null null 3 null null null 8

输出
30,36,21,36,35,26,15,33,8
5.代码

#include <iostream>

#include <queue>

#include <stack>

#include<cstdlib>

#include <cstring>

#include<map>

using namespace std;

struct TreeNode

{
    

	int val;

	TreeNode *left;

	TreeNode *right;

	TreeNode() : val(0), left(NULL), right(NULL) {
    }

	TreeNode(int x) : val(x), left(NULL), right(NULL) {
    }

	TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {
    }

};

TreeNode* inputTree()

{
    

	int n, count = 0;

	char item[100];

	cin >> n;

	if (n == 0)

		return NULL;

	cin >> item;

	TreeNode* root = new TreeNode(atoi(item));

	count++;

	queue<TreeNode*> nodeQueue;

	nodeQueue.push(root);

	while (count < n)

	{
    

		TreeNode* node = nodeQueue.front();

		nodeQueue.pop();

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int leftNumber = atoi(item);

			node->left = new TreeNode(leftNumber);

			nodeQueue.push(node->left);

		}

		if (count == n)

			break;

		cin >> item;

		count++;

		if (strcmp(item, "null") != 0)

		{
    

			int rightNumber = atoi(item);

			node->right = new TreeNode(rightNumber);

			nodeQueue.push(node->right);

		}

	}

	return root;

}

void printTree(TreeNode* root) {
    

	if (root == NULL) {
    

		return;

	}

	bool isFirst = true;

	queue<TreeNode*> q;

	q.push(root);

	while (!q.empty()) {
    

		TreeNode* node = q.front();

		q.pop();

		if (node == NULL) {
    

			continue;

		}

		if (!isFirst)

			cout << ",";

		cout << node->val;

		isFirst = false;

		q.push(node->left);

		q.push(node->right);

	}

}
int sum = 0;
TreeNode*  bstToGst(TreeNode *root)
{
    
	
	//Reverse inorder traversal of a binary search tree
	if (root != NULL)
	{
    
		bstToGst(root->right);//右子树
		sum += root->val;//进行累加操作
		root->val = sum;
		bstToGst(root->left);//左子树
	}
	return root;
}

int main()

{
    

	TreeNode* root;

	root = inputTree();

	TreeNode* res = bstToGst(root);

	printTree(res);

}