LC 889. Construct Binary Tree from Preorder and Postorder Traversal

Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

Example 1:

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Note:

• 1 <= pre.length == post.length <= 30
• pre[] and post[] are both permutations of 1, 2, ..., pre.length.
• It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

Runtime: 20 ms, faster than 10.38% of C++ online submissions for Construct Binary Tree from Preorder and Postorder Traversal.

想用map优化查找left root的速度，结果更慢了....

#include <assert.h>
class Solution {
private:
  unordered_map<int,int>mp;
public:
  TreeNode* constructFromPrePost(vector<int>& pre, vector<int>& post) {
    for(int i=; i<post.size(); i++) mp[post[i]] = i;
    return helper(pre, post, , pre.size()-, , post.size()-);
  }
  TreeNode* helper(vector<int>& pre, vector<int>& post, int pres, int pree, int poss, int pose){
    if(pres > pree || poss > pose) return nullptr;
    assert(pre[pres] == post[pose]);
    int rootval = pre[pres];
    TreeNode* root = new TreeNode(rootval);
    if(pres == pree && poss == pose) return root;
    int leftidx = mp[pre[pres+]];
    int leftlength = leftidx - poss + ;
    int leftpose = pres + leftlength;
    root->left = helper(pre, post, pres+, leftpose, poss, leftidx);
    root->right = helper(pre, post, leftpose+, pree, leftidx+, pose-);
    return root;
  }
};