105. 从前序与中序遍历序列构造二叉树

最新推荐文章于 2025-06-06 21:48:36 发布

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最新推荐文章于 2025-06-06 21:48:36 发布

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文章标签：算法数据结构

本文链接：https://blog.youkuaiyun.com/qq_44902386/article/details/131403963

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该代码实现了一个Solution类，包含两个方法：buildTree用于根据中序遍历和前序遍历构建二叉树，ordercur使用广度优先搜索进行层次遍历。示例中，给定中序遍历和前序遍历的数组，构建出二叉树并输出层次遍历的结果。

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

class Solution {
public:
	TreeNode* buildTree(vector<int>& inorder, vector<int>& preorder) {
		if (preorder.size() == 0)return NULL;
		int value = preorder[0];
		TreeNode* root = new TreeNode(value);
		if (preorder.size() == 1)return root;

		int index = 0;
		for (index = 0; index < inorder.size(); index++) {
			if (inorder[index] == value)
				break;
		}

		vector<int> leftinorder(inorder.begin(),inorder.begin()+index);
		vector<int> rightinorder(inorder.begin() + index+1,inorder.end());

		reverse(preorder.begin(), preorder.end());
		preorder.resize(preorder.size() - 1);
		reverse(preorder.begin(), preorder.end());

		vector<int> leftpreorder(preorder.begin(),preorder.begin()+leftinorder.size());
		vector<int> rightpreorder(preorder.begin() + leftinorder.size(),preorder.end());

		root->left = buildTree(leftinorder, leftpreorder);
		root->right = buildTree(rightinorder, rightpreorder);
		return root;
	}
	vector<int> ordercur(TreeNode* root) {
		queue<TreeNode*> que;
		if (root != NULL) {
			que.push(root);
		}
		vector<int> result;
		while (!que.empty()) {
			int size = que.size();
			for (int i = 0; i < size; i++) {
				TreeNode* node = que.front();
				result.push_back(node->val);
				que.pop();
				if (node->left)que.push(node->left);
				if (node->right)que.push(node->right);
			}
		}
		return result;
	}
};

int main() {

	vector<int> inorder = { 9,3,15,20,7 };
	vector<int> preorder = { 3,9,20,15,7 };
	Solution ss;
	TreeNode* root = ss.buildTree(inorder, preorder);
	vector<int> result = ss.ordercur(root);
	for (vector<int>::iterator pos = result.begin(); pos != result.end(); pos++) {
		cout << *pos << " ";
	}
	cout << endl;

	return 0;
}