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  • 106. Construct Binary Tree from Inorder and Postorder Traversal

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    106. Construct Binary Tree from Inorder and Postorder Traversal

    题目

    Given inorder and postorder traversal of a tree, construct the binary tree.
    
    Note:
    You may assume that duplicates do not exist in the tree. 
    

    解析

    // 106. Construct Binary Tree from Inorder and Postorder Traversal
    class Solution_106 {
    public:
    	TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) { //这样消耗内存多些
    		
    		if (inorder.size()==0||postorder.size()==0||inorder.size()!=postorder.size())
    		{
    			return NULL;
    		}
    
    		int len = postorder.size();
    		TreeNode* root = new TreeNode(postorder[len-1]);
    		
    		//bug:
    		//terminate called after throwing an instance of 'std::bad_alloc'
    		//what() : std::bad_alloc
    
    		auto pos = find(inorder.begin(), inorder.end(), postorder[len - 1]);
    		vector<int> inorder_l(inorder.begin(),pos);
    		vector<int> inorder_r(pos + 1, inorder.end());
    		vector<int> postorder_l(postorder.begin(), postorder.begin() + inorder_l.size());
    		vector<int> postorder_r(postorder.begin()+inorder_l.size(),postorder.end()-1);
    		
    		if (inorder_l.size()>0)
    		{
    			root->left = buildTree(inorder_l, postorder_l);
    		}
    
    		if (inorder_r.size()>0)
    		{
    			root->right = buildTree(inorder_r, postorder_r);
    		}
    		
    		return root;
    	}
    
    
    public:
    	TreeNode* buildTreeHelper(vector<int>& inorder, int l1, int r1, vector<int>& postorder, int l2, int r2) //在原数组上操作,不需要额外空间
    	{
    		if (l1>r1||l2>r2)
    		{
    			return NULL;
    		}
    		
    		TreeNode* root = new TreeNode(postorder[r2]);
    		int i = 0;
    		for ( i= l1; i <= r1;i++) // for ( i= 0; i < inorder.size();i++) //递归实现参数要能进入下次递归
    		{
    			if (inorder[i]==postorder[r2])
    			{
    				break;
    			}
    		}
    
    		root->left = buildTreeHelper(inorder,l1,i-1,postorder,l2,l2+(i-1-l1)); //慢慢体会下标的准确性
    		root->right = buildTreeHelper(inorder, i + 1, r1, postorder, l2 + (i - l1), r2-1);
    
    		return root;
    	}
    
    	TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) 
    	{
    		if (inorder.size()==0||postorder.size()==0||inorder.size()!=postorder.size())
    		{
    			return NULL;
    		}
    
    		return buildTreeHelper(inorder, 0, inorder.size() - 1, postorder,0, postorder.size() - 1);
    	}
    };
    

    题目来源

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  • 原文地址:https://www.cnblogs.com/ranjiewen/p/8259578.html
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