思路:了解了先序遍历和中序遍历的遍历顺序,然后划分左右子树即可。
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
return buildTree(preorder,0,preorder.size()-1,inorder,0,inorder.size()-1);
}
TreeNode* buildTree(vector<int>&preorder,int pleft,int pright,vector<int>&inorder,int ileft,int iright)
{
if(pleft>pright||ileft>iright) return NULL;
int i=0;
for(i=ileft;i<iright;++i)
{
if(preorder[pleft]==inorder[i]) break;
}
TreeNode* cur=new TreeNode(preorder[pleft]);
cur->left=buildTree(preorder,pleft+1,pleft+i-ileft,inorder,ileft,i-1);
cur->right=buildTree(preorder,pleft+i-ileft+1,pright,inorder,i+1,iright);
return cur;
}
};