leetcode -- Construct Binary Tree from Preorder and Inorder Traversal

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

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

 

[解题思路]

由前序遍历知第一个节点是根节点，根据此节点值去中序遍历集合中找该root值所处位置，

该位置之前的数属于左子树，之后的属于右子树，即找到左子树和右子树所处的子序列，接下来就可以用递归来完成

递归的终止条件：preorder中仅有一个元素且preorder中的元素和inorder中元素相同

 1 /**
 2  * @author rgc
 3  * @date 2013-8-23上午10:07:40
 4  * @version 1.0
 5  */
 6 public class Solution {
 7 
 8     public static void main(String[] args) {
 9         int[] preorder = new int[]{1,2,3};
10         int[] inorder = new int[]{1,2,3};
11         buildTree(preorder, inorder);
12     }
13 
14     public static TreeNode buildTree(int[] preorder, int[] inorder) {
15         // Start typing your Java solution below
16         // DO NOT write main() function
17         if (preorder == null || inorder == null) {
18             return null;
19         }
20         int preLen = preorder.length;
21         int inLen = inorder.length;
22         if (preLen == 0 || inLen == 0) {
23             return null;
24         }
25 
26         return constructTree(preorder, 0, preLen - 1, inorder, 0, inLen - 1);
27     }
28 
29     public static TreeNode constructTree(int[] preorder, int preStart, int preEnd,
30             int[] inorder, int inStart, int inEnd) {
31         int rootValue = preorder[preStart];
32         TreeNode root = new TreeNode(rootValue);
33         root.left = null;
34         root.right = null;
35         if (preStart == preEnd && preorder[preStart] == inorder[inStart]) {
36             return root;
37         }
38 
39         int i = inStart;
40         for (; i <= inEnd; i++) {
41             if (rootValue == inorder[i]) {
42                 break;
43             }
44         }
45         int leftLen = i - inStart;
46         // exsit left subtree
47         if (leftLen > 0) {
48             root.left = constructTree(preorder, preStart + 1, preStart
49                     + leftLen, inorder, inStart, i - 1);
50         }
51         if (inEnd > i) {
52             root.right = constructTree(preorder, preStart + leftLen + 1,
53                     preEnd, inorder, i + 1, inEnd);
54         }
55         return root;
56     }
57 
58 }
59 
60 class TreeNode {
61     int val;
62     TreeNode left;
63     TreeNode right;
64 
65     TreeNode(int x) {
66         val = x;
67     }
68 }