Leetcode | | 106. Construct Binary Tree from inorder and Postorder traversal

Problem:

Given Inorder and Postorder traversal of a tree, construct the binary tree.

Note:
Assume that duplicates does not exist in the tree.

Hide TagsTree Array Depth-first SearchTest instructions: The binary tree is constructed by the sequential traversal sequence of the given binary tree and the subsequent traversal sequence.

Thinking:

(1) This type of problem, to give an example to find out its law. For a full two-fork tree (1,2,3,4,5,6,7), the middle sequence traversal: 4,2,5,1,6,3,7 subsequent traversal: 4,5,2,6,7,3,1

First, the root node 1 is in the last position of the subsequent traversal sequence, and then in the middle sequence to traverse the position of Find 1, before 1 is the left dial hand tree, and after 1 is the right subtree. Once you know the number of child tree nodes,

It is also possible to divide the left and right sub-trees for subsequent traversal sequences.

(2) Repeating the above procedure recursively

Code

Class Solution {public  :      TreeNode *buildtree (vector<int> &inorder, Vector<int> & Postorder) {          if (inorder.size () = = 0)              return NULL;          Return make (Inorder.begin (), Inorder.end (), Postorder.begin (), Postorder.end ());          }  Protected:      template<class it>      TreeNode *make (it pfirst,it plast,it qfirst,it qlast)      {          if ( Pfirst==plast)              return NULL;          it loc1 = qLast-1;          int a = *loc1;          It Loc2=find (pfirst,plast,a);          int Left_size=loc2-pfirst;          TreeNode *root=new TreeNode (a);          Root->left=make (pfirst,loc2,qfirst,qfirst+left_size);          Root->right=make (loc2+1,plast,qfirst+left_size,qlast-1);          return root;      }  };

Leetcode | | 106. Construct Binary Tree from inorder and Postorder traversal