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

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

利用中序和後序遍歷構造二叉樹，要註意到後序遍歷的最後一個元素是二叉樹的根節點，而中序遍歷中，根節點前面為左子樹節點後面為右子樹的節點。例如二叉樹：{1,2,3,4,5,6,#}的後序遍歷為4->5->2->6->3->1；中序遍歷為：4->2->5->1->6->3。

故，遞歸的整體思路是，找到根節點，然後遍歷左右子樹。

方法一：遞歸

值得註意的是：遞歸過程中，起始點的選取。Grandyang對下標的選取有較為詳細的說明。

 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：

利用棧。這個方法是以前看到出處也尷尬的忘了，分析過程等我再次看懂了再補上。這種方法改變了數組。

 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 };

[Leetcode] Construct binary tree from inorder and postorder travesal 利用中序和後續遍歷構造二叉樹