105. Construct Binary Tree from Preorder and Inorder Traversal

Posted 2020-10-20 小河沟大河沟

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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;
    }

};

题目来源

• 105. Construct Binary Tree from Preorder and Inorder Traversal