[leetcode]Construct Binary Tree from Preorder and Inorder Traversal

從前序中序遍歷來重構二叉樹

經典題,從期末考試到考研什么的應該都有

 

前序遍歷第一個肯定是root

在inorder里面去找root

左邊的是leftsubtree , 右邊的是rightsubtree

然后preorder除去order,后面lifetsubtree.size 個是左邊的, rightsubtree.size個是右邊的

遞歸構造tree就好了

 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    template<typename Iter>
    TreeNode* make(Iter pFirst , Iter pLast , Iter iFirst , Iter iLast) {
        if(pFirst == pLast) return nullptr;
        if(iFirst == iLast) return nullptr;
        int val = *pFirst;
        auto iRoot = find(iFirst , iLast , val);
        TreeNode* root = new TreeNode(*iRoot);
        int leftSize = iRoot - iFirst;
        root -> left = make(pFirst+1 , pFirst+leftSize+1 , iFirst , iRoot);
        root -> right = make(pFirst+leftSize+1 , pLast , iRoot + 1 , iLast);
        return root;
    }
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        int size = inorder.size();
        if(size == 0) return nullptr;
        return make(preorder.begin() , preorder.end() , inorder.begin() , inorder.end());
    }
};