LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal

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

题目：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

题解：

这道题与Construct Binary Tree from Preorder and Inorder Traversal思路相似，不同之处在于这里的root在postorder的最有一位，其他都相同，由inorder找出分切点.

Time Complexity: O(n). Space: O(n).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode buildTree(int[] inorder, int[] postorder) {
12         if(inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0){
13             return null;
14         }
15         HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>();
16         for(int i = 0; i<inorder.length; i++){
17             hm.put(inorder[i], i);
18         }
19         return helper(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1, hm);
20     }
21     
22     private TreeNode helper(int [] inorder, int inL, int inR, int [] postorder, int posL, int posR, HashMap<Integer, Integer> hm){
23         if(inL > inR || posL > posR){
24             return null;
25         }
26         TreeNode root = new TreeNode(postorder[posR]);
27         int rootIndex = hm.get(postorder[posR]);
28         root.left = helper(inorder, inL, rootIndex-1, postorder, posL, posL+rootIndex-inL-1, hm);
29         root.right = helper(inorder, rootIndex+1, inR, postorder, posL+rootIndex-inL, posR-1, hm);
30         return root;
31     }
32 }

类似Construct Binary Tree from Preorder and Postorder Traversal.

posted @ 2015-08-21 03:03  Dylan_Java_NYC  阅读(218)  评论(0)  编辑  收藏  举报