学步园

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

Note:

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

TreeNode* build(vector<int> &preorder,int left1,int right1,vector<int> &inorder,int left2,int right2)
	{
		if(right1-left1 != right2-left2)return NULL;
		if(right1>=preorder.size() || right2>=inorder.size())return NULL;

		if(left1==right1 && left2==right2)
		{
			TreeNode* root = new TreeNode(preorder[left1]);
			return root;
		}else if(left1<right1 && left2<right2)
		{
			TreeNode* root = new TreeNode(preorder[left1]);
			int i;
			for(i = left2; i <= right2; i++)
				if(inorder[i] == preorder[left1])break;
			if(i>right2)return NULL;
			root->left = build(preorder,left1+1,left1+i-left2,inorder,left2,i-1);
			root->right = build(preorder,left1+i-left2+1,right1,inorder,i+1,right2);
			return root;
		}else
			return NULL;
	}
	TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // Note: The Solution object is instantiated only once.
        return build(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
    }