LeetCode 889. Construct Binary Tree from Preorder and Postorder Traversal(Medium)

Description:
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.

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Analysis:
The key point of the problem is to figure out the intervals of indexes of left subtree and right subtree in pre  and post respectively. Here we use L to denote the number of nodes in the left subtree. Then we can get the following graph:

As the following graph show, traverse the post  until an element that is equal to pre[1]is found, this element represents the root node of the left subtree in post, then we can obtain the value of L. We can use a recursion to apply this strategy to construct the root node, left subtree and right subtree.

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Code:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode constructFromPrePost(int[] pre, int[] post) {
        if(pre.length == 0) {
            return null;
        }else if(pre.length == 1) {
            return new TreeNode(pre[0]);
        }else{
            TreeNode root = new TreeNode(pre[0]);
            
            int N = pre.length;
            int L = 0;
            for(int i = 0; i < N; i++) {
                L += 1;
                if(post[i] == pre[1]) {
                    break;
                }
            }
            
            root.left = constructFromPrePost(Arrays.copyOfRange(pre, 1, L+1), Arrays.copyOfRange(post, 0, L));
            root.right = constructFromPrePost(Arrays.copyOfRange(pre, L+1, N), Arrays.copyOfRange(post, L, N-1));
            
            return root;
        }
    }
}