functionclass[LeetCode]Construct Binary Tree from Preorder and Inorder Traversal

废话就不多说了，开始。。。

    每日一道理
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struct TreeNode {
	int val;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

class Solution {
	//Note:
	//You may assume that duplicates do not exist in the tree.
	//so we can build an unordered_map O(1) to look up the index in inorder, to divide the left subtree and right subtree
	//in postorder.
	//inorder{(inStart)left(inRootIdx-1)root(inRootIdx+1)right}, postorder{left(inRootIdx-1)right(PostEnd-1)root}
public:
	unordered_map<int, int> m_Value2Index;//inorder map
	void BuildMap(vector<int> &inorder)
	{
		m_Value2Index.clear();
		for(int i = 0; i < inorder.size(); ++i)
			m_Value2Index[inorder[i]] = i;
	}
	//just make sure (inHigh-inLow == preHigh-preLow), there will be no problem then
	TreeNode*  BuildTreeInPlusPre(vector<int> &inorder, int inLow, int inHigh, vector<int> &preorder, int preLow, int preHigh )
	{
		if(inLow > inHigh || preLow > preHigh)
			return NULL;
		int rootValue = preorder[preLow];
		TreeNode* parent = new TreeNode(rootValue);
		int inRootIdx = m_Value2Index[rootValue];
		parent->left = BuildTreeInPlusPre(inorder, inLow, inRootIdx-1, preorder, preLow+1, preLow+inRootIdx-inLow);
		parent->right = BuildTreeInPlusPre(inorder, inRootIdx+1, inHigh, preorder, preHigh-(inHigh-inRootIdx-1), preHigh);
		return parent;
	}
	TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		BuildMap(inorder);
		return BuildTreeInPlusPre(inorder, 0, inorder.size()-1, preorder, 0, preorder.size()-1);
	}

};

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function和class
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