LeetCode105 Construct Binary Tree from Preorder and Inorder Traversal 通过前序和中序遍历构造二叉树

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

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

题源：here；完整实现：here

思路：

通过递归的方式构造二叉树。两种实现方式，最后解释差异：

1 直接拷贝

TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
	if (preorder.size() == 0) return NULL;
	int val = preorder[0];

	int left = 0;
	while (inorder[left] != val) left++;
	TreeNode *root = new TreeNode(val);
	vector<int> preorderLeft(preorder.begin()+1, preorder.begin() + left+1);
	vector<int> inorderLeft(inorder.begin(), inorder.begin() + left);
	root->left = buildTree(preorderLeft, inorderLeft);
	vector<int> preorderRight(preorder.begin() + left+1, preorder.end());
	vector<int> inorderRight(inorder.begin() + left+1, inorder.end());
	root->right = buildTree(preorderRight, inorderRight);
		
	return root;
}

 2 跟踪标号

TreeNode* buildTree2(vector<int>& preorder, vector<int>& inorder, vector<int> p){
	if (p[0] < 0 || p[2] < 0 || p[1] < p[0] || p[3] < p[2] || p[0] >= preorder.size() || p[2] >= inorder.size())
		return NULL;
	int val = preorder[p[0]];
	int left = p[2];
	while (inorder[left] != val) left++;
	TreeNode *root = new TreeNode(val);
	vector<int> pLeft = p;
	pLeft[0]++;
	pLeft[1] = p[0]+left-p[2];
	pLeft[3] = left-1;
	root->left = buildTree2(preorder, inorder, pLeft);
	vector<int> pRight = p;
	pRight[0] += left-p[2] + 1;
	pRight[2] = left + 1;
	root->right = buildTree2(preorder, inorder, pRight);

	return root;
}

TreeNode* buildTree2(vector<int>& preorder, vector<int>& inorder){
	vector<int> p;
	p.push_back(0);
	p.push_back(preorder.size() - 1);
	p.push_back(0);
	p.push_back(inorder.size() - 1);
	return buildTree2(preorder, inorder, p);
}

可以看到，第一种方式简单直观，但是中间有很多赋值操作，会大量增加运行时间和内存使用，第二种使用索引的方式来构造二叉树，虽然逻辑繁琐一点（我已经晕了，所以第一个判断我就快刀斩乱麻，想到什么加什么了），但是运行时间较快，消耗内存少。