LeetCode OJ：Construct Binary Tree from Inorder and Postorder Traversal（从中序以及后序遍历结果中构造二叉树）

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

Note:
You may assume that duplicates do not exist in the tree.
同样是给出了不会有重复数字的条件，用递归较容易实现，代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
13         if(inorder.size() == 0)
14             return NULL;
15         return createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);   
16     }
17 
18     TreeNode* createTree(vector<int> & inorder, int inBegin, int inEnd,
19      vector<int> & postorder, int postBegin, int postEnd)
20     {
21         if(inBegin > inEnd) return NULL;
22         int rootVal = postorder[postEnd];
23         int mid;
24         for(int i = inBegin; i <= inEnd; ++i){
25             if(inorder[i] == rootVal){
26                 mid = i;
27                 break;
28             }
29         }
30         int len = mid - inBegin;
31         TreeNode * left = createTree(inorder, inBegin, mid - 1,     //边界条件同样应该注意
32                             postorder, postBegin, postBegin + len - 1);
33         TreeNode * right = createTree(inorder, mid + 1, inEnd,
34                             postorder, postBegin + len, postEnd - 1);
35         TreeNode * root = new TreeNode(rootVal);
36         root->left = left;
37         root->right = right;
38         return root;
39     }
40 };

 java版本的代码如下所示，没有区别：

 1 public class Solution {
 2     public TreeNode buildTree(int[] inorder, int[] postorder) {
 3         if(inorder.length == 0)
 4             return null;
 5         return createTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);       
 6     }
 7     
 8     public TreeNode createTree(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd){
 9         if(inEnd < inBegin)
10             return null;
11         int rootVal = postorder[postEnd];
12         int mid = 0;
13         for(int i = inBegin; i <= inEnd; ++i){
14             if(inorder[i] == rootVal){
15                 mid = i;
16                 break;   
17             }
18         }
19         int len = mid - inBegin;
20         TreeNode leftNode = createTree(inorder, inBegin, mid - 1, postorder, postBegin, postBegin + len - 1);
21         TreeNode rightNode = createTree(inorder, mid + 1, inEnd, postorder, postBegin + len, postEnd - 1);
22         TreeNode root = new TreeNode(rootVal);
23         root.left = leftNode;
24         root.right = rightNode;
25         return root;        
26     }
27 }

 

posted @ 2015-10-26 09:46  eversliver  阅读(195)  评论(0)  编辑  收藏  举报