二叉树重建（前+中，后+中，层+中）_二叉树的重建(前序+中序)(后序+中序)(层序+中序)-优快云博客

本文链接：https://blog.youkuaiyun.com/yuyanhaonan/article/details/124340712

本文介绍了如何使用递归构建二叉搜索树，并通过中序遍历实现节点组织。首先，展示了使用中序元素创建节点的结构，然后分别讲解了两段代码，一是利用层序序列构建二叉树，二是根据中序序列构建。重点在于理解树的结构和遍历方式的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

前+中

struct node {
    int num;
    node* l, * r;
    node(int elem, node* n1 = nullptr, node* n2 = nullptr) {
        num = elem;
        l = n1, r = n2;
    }
};

int mei[20], pre[20];

node* build(int lp,int rp,int lm,int rm) {
    if (lm > rm || lp > rp) return nullptr;
    int id = 0,ls;
    for (int i = lm; i <= rm; ++i) {
        if (mei[i] == pre[lp]) {
            id = i;
            break;
        }
    }
    ls = id - lm;
    node* tree= new node(pre[lp]);
    tree->l = build(lp+1,lp+ls,lm,id-1);
    tree->r = build(lp+ls+1,rp,id+1,rm);
    return tree;
}

后+中

struct node {
    int num;
    node* l, * r;
    node(int elem, node* n1 = nullptr, node* n2 = nullptr) {
        num = elem;
        l = n1, r = n2;
    }
};

int mei[20], beh[20];

node* build(int lm,int rm,int lb,int rb) {
    if (lm > rm || lb > rb) return nullptr;
    int id = 0,ls;
    for (int i = lm; i <= rm; ++i) {
        if (mei[i] == beh[rb]) {
            id = i;
            break;
        }
    }
    ls = id - lm;
    node* tree= new node(beh[rb]);
    tree->l = build(lm,id-1,lb,lb+ls-1);
    tree->r = build(id+1,rm,lb+ls,rb-1);
    return tree;
}

层+中

struct Node {
	int data;
	Node* lchild;
	Node* rchild;
	Node(const int& elem, Node* lc = nullptr, Node* rc = nullptr) {
		data = elem;
		lchild = lc;
		rchild = rc;
	}
};

int* sqe;//层序
int* med;//中序

Node*  buildTree(int sL, int sR, int mL, int mR) {
	if (mL > mR) return nullptr;

	int i, j;
	int flag = 0;
	for (i = sL; i <= sR; i++) {//在层序中找到第一个在中序中存在的数字
		for (j = mL; j <= mR; j++) {
			if (sqe[i] == med[j]) {
				flag = 1;
				break;
			}
		}
		if (flag) break;
	}
	
	Node* root = new Node(sqe[i]);
	root->lchild = buildTree(sL + 1, sR, mL, j - 1);
	root->rchild = buildTree(sL + 1, sR, j + 1, mR);

	return root;
}