[Leetcode] Construct binary tree from inorder and postorder travesal 利用中序和后续遍历构造二叉树

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

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

利用中序和后序遍历构造二叉树，要注意到后序遍历的最后一个元素是二叉树的根节点，而中序遍历中，根节点前面为左子树节点后面为右子树的节点。例如二叉树：{1,2,3,4,5,6,#}的后序遍历为4->5->2->6->3->1；中序遍历为：4->2->5->1->6->3。

故，递归的整体思路是，找到根节点，然后遍历左右子树。

方法一：递归

值得注意的是：递归过程中，起始点的选取。Grandyang对下标的选取有较为详细的说明。

 1 /**
 2  * Definition for binary tree
 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     {
14         int len=inorder.size();
15         return build(inorder,0,len-1,postorder,0,len-1);
16     }
17     TreeNode *build(vector<int> &inorder,int inBeg,int inEnd,vector<int> &postorder,int postBeg,int postEnd)
18     {
19         if(inBeg>inEnd||postBeg>postEnd)    return NULL;
20         TreeNode *root=new TreeNode(postorder[postEnd]);
21
22         for(int i=0;i<postorder.size();++i)
23         {
24             if(inorder[i]==postorder[postEnd])
25             {
26                 root->left=build(inorder,inBeg,i-1,postorder,postBeg,postBeg+i-1-inBeg);
27                 root->right=build(inorder,i+1,inEnd,postorder,postBeg+i-inBeg,postEnd-1);
28             }
29         }
30         return root;
31     }
32 };

方法二：

利用栈。这个方法是以前看到出处也尴尬的忘了，分析过程等我再次看懂了再补上。这种方法改变了数组。

 1 /**
 2  * Definition for binary tree
 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     {
14         if(inorder.size()==0)   return NULL;
15         TreeNode* p;
16         TreeNode* root;
17         stack<TreeNode*> stn;
18
19         root=new TreeNode(postorder.back());
20         stn.push(root);
21         postorder.pop_back();
22
23         while(true)
24         {
25             if(inorder.back()==stn.top()->val)
26             {
27                 p=stn.top();
28                 stn.pop();
29                 inorder.pop_back();
30
31                 if(inorder.size()==0)   break;
32                 if(stn.size()&&inorder.back()==stn.top()->val)
33                     continue;
34
35                 p->left=new TreeNode(postorder.back());
36                 postorder.pop_back();
37                 stn.push(p->left);
38             }
39             else
40             {
41                 p=new TreeNode(postorder.back());
42                 postorder.pop_back();
43                 stn.top()->right=p;
44                 stn.push(p);
45             }
46         }
47         return root;
48     }
49 };