DFS,通过preorder和inorder得到树。
/**
* 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* buildTree(int i1,int j1,int i2,int j2,vector<int>& preorder, vector<int>& inorder)
{
if(i1==j1&&i2==j2)
{
TreeNode* root=new TreeNode(preorder[i1]);
return root;
}
else if(i1>j1||i2>j2)
return NULL;
else
{
TreeNode* root=new TreeNode(preorder[i1]);
int mid;
for(mid=i2;mid<=j2;mid++)
{
if(preorder[i1]==inorder[mid])
break;
}
//mid---i1
root->left=buildTree(i1+1,i1+mid-i2,i2,mid-1,preorder, inorder);
root->right=buildTree(i1+mid-i2+1,j1,mid+1,j2,preorder, inorder);
return root;
}
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
if(preorder.size()==0)
return NULL;
return buildTree(0,preorder.size()-1,0,inorder.size()-1,preorder,inorder);
}
};