LeetCode 889. Construct Binary Tree from Preorder and Postorder Traversal

原题链接在这里：https://leetcode.com/problems/construct-binary-tree-from-preorder-and-postorder-traversal/

题目：

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.

题解：

The first element in pre, 1 is actually the root's val.

The second element in pre, 2 is root's left child val. Find 2's index in post, before that, all are root left subtree.

Vice Versa.

Use a HashMap to store post element and its index for quick search.

In recursion, before using pre[preL+1], be careful with OutOfBoundException. Return when preL == preR. This makes sure after that, preL will not be out of index bound.

Time Complexity: O(n).

Space: O(n).

AC Java:

 /**
  * 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 == null || pre.length == 0 || post == null || post.length == 0){
             return null;
         }

         HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
         for(int i = 0; i<post.length; i++){
             hm.put(post[i], i);
         }

         return construct(pre, 0, pre.length-1, post, 0, post.length-1, hm);
     }

     private TreeNode construct(int[] pre, int preL, int preR, int[] post, int postL, int postR, HashMap<Integer, Integer> hm){
         if(preL > preR){
             return null;
         }

         TreeNode root = new TreeNode(pre[preL]);
         if(preL == preR){
             return root;
         }

         int leftVal = pre[preL+1];
         int leftIndex = hm.get(leftVal);
         root.left = construct(pre, preL+1, preL+1+leftIndex-postL, post, postL, leftIndex, hm);
         root.right = construct(pre, preL+2+leftIndex-postL, preR, post, leftIndex+1, postR-1, hm);
         return root;
     }
 }

类似Construct Binary Tree from Preorder and Inorder Traversal, Construct Binary Tree from Inorder and Postorder Traversal.