Rebuild Any Binary Tree Matching Preorder and Postorder

### Problem Given preorder and postorder traversals of the same binary tree with distinct values, construct and return a tree matching both traversals. If several binary-tree shapes are consistent, any one is accepted. Preserve the native TreeNode return interface; level-order arrays below only display example trees. ```cpp class Solution { public: TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder); }; ``` ### Examples **Example 1:** ```text Input: preorder = [2,1,3], postorder = [1,3,2] Output: tree displayed as[2,1,3] Explanation: Root2 has left child1 and right child3, matching both traversals. ``` **Example 2:** ```text Input: preorder = [1,2], postorder = [2,1] Output: tree displayed as[1,2] Explanation: Making2 the left child of1 works. Making it the right child also gives the same traversals. ``` ### Constraints - `1 <= preorder.length <= 30`. - `postorder.length == preorder.length`. - Every value in either array lies between1 and preorder.length, inclusive. - Values in each traversal are unique. - Both arrays are guaranteed traversals of the same binary tree. ### Contract considerations Traversals are guaranteed mutually valid and use distinct values. Return a TreeNode pointer, not an invented array wrapper. ```hint Read the boundary condition Locate the next preorder root in postorder to find the first child subtree size. ``` ### What a Strong Answer Covers Traversals are guaranteed mutually valid and use distinct values. Return a TreeNode pointer, not an invented array wrapper. ### Follow-up Questions Optional practice extension, separate from the original task: What additional traversal information would make every valid tree reconstruction unique?