Leetcode 106 Construct Binary Tree from Inorder and Postorder Traversal

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

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

中序和后序还原二叉树。

直接在上一题的代码上做一点小修改就行了。

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* dfs(vector<int>& preorder, int ps,int pe,vector<int>& inorder,int is,int ie)      {          if(ps>pe ) return NULL;          TreeNode* root=new TreeNode(preorder[pe]);          int pos;          for(int i=is;;i++)          if(inorder[i]==preorder[pe])          {              pos=ie-i;              break;          }          root->left=dfs(preorder,ps,pe-pos-1,inorder,is,ie-pos-1);          root->right=dfs(preorder,pe-pos,pe-1,inorder,ie-pos+1,ie);          return root;      }      TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {        return dfs(postorder,0,inorder.size()-1,inorder,0,inorder.size()-1);    }};