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  • [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。这里有一个很重要的条件是:树中没有重复元素。

    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。至于这个递归过程,(个人愚见,欢迎大神给出更好递归过程)可以在脑海想,构造顺序为4->5->2构建好左子树,接上1,然后6->3构建好右子树,最后接上1,完成。

     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<inorder.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 };
    View Code
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  • 原文地址:https://www.cnblogs.com/love-yh/p/6963736.html
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