Leetcode---Construct Binary Tree from Preorder and Inorder Traversal

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

由树的前序遍历序列和中序遍历序列构建二叉树

1. 在中序序列中找到根节点

2. 把中序序列拆分成左右两个中序序列

3. 找对应的前序序列，注意，对应的中序和前序序列长度相同

4. 递归

TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
    if(preorder.size()==0 || inorder.size()==0)
        return NULL;
    TreeNode * root=new TreeNode(preorder[0]);
    int j=0;
    for(j=0;j<inorder.size();j++){
        if(inorder[j]==preorder[0]){
            break;
        }
    }
    if(j-1>=0){
    vector<int> leftinorder(inorder.begin(),inorder.begin()+j-1);
    vector<int> leftpreorder(preorder.begin()+1,preorder.begin()+j);
    root->left=buildTree(leftpreorder,leftinorder);
    }
    if(j+1<=inorder.size()-1){
    vector<int> rightinorder(inorder.begin()+j+1,inorder.end());
    vector<int> rightpreorder(preorder.begin()+1+j,preorder.end());
    root->right=buildTree(rightpreorder,rightinorder);
    }
    return root;
}

哈哈，超内存了！原因就是局部变量用多了呗，修改一下：

TreeNode * f(vector<int> &inorder, vector<int> &preorder, int inl,int inr,int prel, int prer){
    int n=inr-inl+1;
    if(n==0)
        return NULL;
    int i=0;
    for(;i<=n-1;i++){
        if(inorder[inl+i]==preorder[prel])
            break;
    }
    TreeNode * root=new TreeNode(inorder[inl+i]);
    root->left=f(inorder,preorder,inl,inl+i-1,prel+1,prel+i);
    root->right=f(inorder,preorder,inl+i+1,inr,prel+i+1,prer);
    return root;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
    return f(inorder,preorder,0,inorder.size()-1,0,preorder.size()-1);
}