题目描述

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

 注意：举例子进行编程，注意边界情况，还有就是注意不要忘记递归基。

不要就举三个节点的例子，可以举七个节点的例子，得到前序和中序遍历。

4,3,1,2,5,6,7

1,3,2,4,6,5,7

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* helper(vector
     
       &pre,int preBegin,int preEnd,vector
      
        &vin,int vinBegin,int vinEnd){        if(preBegin > preEnd || vinBegin > vinEnd){            return nullptr;        }        TreeNode* root = new TreeNode(pre[preBegin]);        //find left        int i = 0;        for(i = vinBegin;i <= vinEnd;++i){            if(vin[i] == root -> val){                break;            }        }        root -> left = helper(pre,preBegin + 1,preBegin + i - vinBegin,vin,vinBegin,i - 1);        root -> right = helper(pre,preBegin + i - vinBegin + 1,preEnd,vin,i + 1,vinEnd);        return root;    }        TreeNode* reConstructBinaryTree(vector
       
         pre,vector
        
          vin) {        if(pre.size() == 0){            return nullptr;        }        TreeNode* root = helper(pre,0,pre.size() - 1,vin,0,vin.size() - 1);        return root;    }};
        
       
      
     

 

 

拷贝数组的效率太低，不要看了。

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    TreeNode* reConstructBinaryTree(vector
     
       pre,vector
      
        vin) {        if(pre.size() != vin.size()){            return nullptr;        }        if(pre.size() == 0){
   //递归基            return nullptr;        }        TreeNode* root = new TreeNode(pre[0]);        vector
       
         vin_left,vin_right,pre_left,pre_right;        int pos = 0;        for(pos;pos < vin.size();++pos){            if(vin[pos] == root -> val){                break;            }            vin_left.push_back(vin[pos]);            pre_left.push_back(pre[pos + 1]);        }        pos++;        for(pos;pos < vin.size();++pos){                        vin_right.push_back(vin[pos]);            pre_right.push_back(pre[pos]);        }                root -> left = reConstructBinaryTree(pre_left,vin_left);        root -> right = reConstructBinaryTree(pre_right,vin_right);        return root;    }};