<Prove safety offer> question 4

topic:

And enter the result in a preorder traversal of a binary tree in preorder traversal of the binary tree is reconstructed requested. Suppose Results preorder traversal of the input sequence and are free of duplicate numbers.

E.g:

Preorder traversal sequence {1,2,4,7,3,5,6,8}

Inorder traversal sequence {4,7,2,1,5,3,8,6}

And outputs the reconstructed binary its head node

 

Ideas:

1. The first node traversed by the root is a known preamble, whereby the preorder divided into two parts, the left part of the left subtree of the root node {4,7,2} (inOrder), the right part of the right subtree of the root node {5,3,8,6} (inOrder)

2. ago by {4,7,2} (inOrder) and Comparative know preorder traversal order traversal is {2,4,7} (preOrder)

　2 can be seen to the left subtree of the first stage "root", which is a left subtree {4,7} (inOrder), which is empty right subtree, which traverse the preamble is {4,7} (preOrder)

　4 can be seen to the left of the second stage of the sub-tree "root", which is left empty subtree, right subtree is {7}

　This, the left sub-tree traversal is complete, right subtree empathy

3. The above ideas recursion

 

Code

public class Third {
   class TreeNode{
       int val;
       TreeNode left;
       TreeNode right;
       public TreeNode(int val){
           this.val = val;
       }
   }

   public void reConstructBST(int[] pre, int[] in){
        reBuildBST(pre, 0, pre.length - 1, in, 0, in.length - 1);
   }

   private TreeNode reBuildBST(int[] pre, int preStart, int preEnd, int[] in, int inStart, int inEnd){
       if(preStart > preEnd || inStart > inEnd){
           return null;
       }
       TreeNode root = new TreeNode(pre[preStart]);
       for(int i = 0; i < inEnd; i ++){
           if(in[i] == pre[preStart]){
               root.left = reBuildBST(pre, preStart, preStart + i - inStart, in, inStart, i - 1);
               root.right = reBuildBST(pre, preStart + i - inStart + 1, preEnd, in, i + 1, inEnd);
           }
       }
       return root;
   }
}