105. Construct Binary Tree from Preorder and Inorder Traversal

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105. Construct Binary Tree from Preorder and Inorder Traversal

题目

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree. 

解析

// Construct Binary Tree from Preorder and Inorder Traversal
class Solution_105 {
public:

	//运行时间：9ms
	//占用内存：640k

	TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {

		if (preorder.size()==0||inorder.size()==0||preorder.size()!=inorder.size())
		{
			return NULL;
		}

		TreeNode* root = new TreeNode(preorder[0]);

		if (preorder.size() == inorder.size() && inorder.size() == 1)
		{
			return root;
		}

		//auto pos = inorder.size() > 1 ? find(inorder.begin(), inorder.end(), preorder[0]) : inorder.begin();
		auto pos = find(inorder.begin(), inorder.end(), preorder[0]) ; 

		//preorder用下标分开也可以
		vector<int> inorder1(inorder.begin(), pos);  //pos指向容器最后一个元素的下一个位置
		vector<int> inorder2(pos + 1, inorder.end());

		vector<int> preorder1(preorder.begin() + 1, preorder.begin() + 1 + inorder1.size()); //cnt=inorder1.size()
		vector<int> preorder2(preorder.begin() + 1 + inorder1.size(), preorder.end());


		//auto iter = preorder.begin();
		//int cnt = 0;
		//while (iter != pos)
		//{
		//	iter++;
		//	cnt++;
		//}

		////preorder用下标分开也可以
		//vector<int> inorder1(inorder.begin(), pos);
		//vector<int> inorder2(pos + 1, inorder.end());
		//vector<int> preorder1(preorder.begin() + 1, preorder.begin() + 1 + cnt); //cnt=inorder1.size() //报alloc错误
		//vector<int> preorder2(preorder.begin() + 1 + cnt, preorder.end());

		if (preorder1.size()>0)
		{
			root->left = buildTree(preorder1, inorder1);
		}
		
		if (preorder2.size()>0)
		{
			root->right = buildTree(preorder2, inorder2);
		}
		
		return root;
	}


	TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
		return build(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1);
	}
	TreeNode *build(vector<int> &preorder, vector<int> &inorder, int l1, int r1, int l2, int r2)
	{
		if (l1 > r1)
			return NULL;
		int gen = preorder[l1];
		int i, cnt = 0;

		for (i = l2; i <= r2&&inorder[i] != gen; cnt++, i++); //找到当前根节点在inorder中的位置

		TreeNode *root = (TreeNode *)malloc(sizeof(TreeNode));
		root->val = gen;
		root->left = build(preorder, inorder, l1 + 1, l1 + cnt, l2, i - 1); //位置信息要准确
		root->right = build(preorder, inorder, l1 + 1 + cnt, r1, i + 1, r2);
		return root;
	}


public:
	using iter = std::vector<int>::iterator;
public:
	TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder)
	{
		return buildTreeHelper(inorder.begin(), inorder.end(), postorder.begin(), postorder.end());
	}

	TreeNode* buildTreeHelper(iter inOrderBegin, iter inOrderEnd, iter postOrderBegin, iter postOrderEnd)
	{
		if (inOrderBegin == inOrderEnd)
			return nullptr;
		if (std::next(inOrderBegin) == inOrderEnd)
			return new TreeNode(*inOrderBegin);
		TreeNode *root = new TreeNode(*std::prev(postOrderEnd));
		auto pivot = std::find(inOrderBegin, inOrderEnd, root->val);
		auto leftSize = std::distance(inOrderBegin, pivot);
		auto rightSize = std::distance(pivot, inOrderEnd) - 1;
		if (leftSize != 0)
			root->left = buildTreeHelper(inOrderBegin, pivot, postOrderBegin, std::next(postOrderBegin, leftSize));
		if (rightSize != 0)
			root->right = buildTreeHelper(std::next(pivot), inOrderEnd, std::next(postOrderBegin, leftSize), std::prev(postOrderEnd));
		return root;
	}
};

题目来源

• 105. Construct Binary Tree from Preorder and Inorder Traversal