題目：

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.

解釋：
按照前序遍歷和後序遍歷的結果恢復二叉樹，注意前序遍歷和後序遍歷並不能唯一地確定一顆二叉樹。
python程式碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
 

#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def constructFromPrePost(self, pre, post):
        """
        :type pre: List[int]
        :type post: List[int]
        :rtype: TreeNode
        """
        root_val=pre[0]
        root=TreeNode(
 
root_val)
        if len(pre) == 1:
            return root
        index=post.index(pre[1])
        pre_left,post_left=pre[1:index+2],post[:index+1]
        pre_right,post_right=pre[index+2:],post[index+1:-1]
        
        if pre_left:
            root.left=self.constructFromPrePost(pre_left,post_left)
        if pre_right:
            root.right=self.constructFromPrePost(pre_right,post_right)
        return root

c++程式碼：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* constructFromPrePost(vector<int>& pre, vector<int>& post) {
        int root_val=pre[0];
        TreeNode* root=new TreeNode(root_val);
        if(pre.size()==1)
            return root;
        int idx=find(post.begin(),post.end(),pre[1])-post.begin();
        vector<int> pre_left(pre.begin()+1,pre.begin()+idx+2);
        vector<int> post_left(post.begin(),post.begin()+idx+1);
        vector<int> pre_right(pre.begin()+idx+2,pre.end());
        vector<int> post_right(post.begin()+idx+1,post.end()-1);
        if(pre_left.size())
            root->left=constructFromPrePost(pre_left, post_left);
        if(pre_right.size())
            root->right=constructFromPrePost(pre_right, post_right);
        return root;
    }
};

總結：
dfs一定要學會在合適的地方判斷遞迴條件，判斷的位置合適的話，可以節省大量的時間。