Construct Binary Tree from Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
一个小小的下标没搞好,居然要浪费差不多一个小时,心痛啊。
中序遍历序列和后序遍历序列的偏移位置都需要仔细设计好。
class Solution {
public:
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder)
{
return conTree(preorder, inorder, 0, preorder.size()-1, 0, inorder.size()-1);
}
TreeNode *conTree(vector<int> &preo, vector<int> &ino,
int prel, int prer, int inl, int inr)
{
if (prel > prer) return NULL;
TreeNode *t = new TreeNode(preo[prel]);
auto it = find(ino.begin()+inl, ino.begin()+inr+1, preo[prel]);
int offset = it-ino.begin()-inl;
//注意:切记,下标如何定位要清楚,是offset还是绝对位置不能搞混了!
t->left = conTree(preo, ino, prel+1, prel+offset, inl, inl+offset-1);
t->right = conTree(preo, ino, prel+offset+1, prer, inl+offset+1, inr);
return t;
}
};