笔试程序题专项----二叉树中两个节点的最低公共祖先节点和距离

//求两个结点的最低公共祖先最低公共祖先，即 LCA（Lowest Common Ancestor）//
//复杂性和 n 同样数量级 
BTree FindLCA(BTree T, BTree target1, BTree target2)
{
    if (T == NULL)
        return NULL;
    //cout<<"vist"<<T->data<<endl;    
    if (T == target1 || T == target2)
        return T;
  
    BTree left = FindLCA(T->lchild, target1, target2);
    BTree right = FindLCA(T->rchild, target1, target2);
    if (left && right)  // 分别在左右子树
        return T;
  
    return left ? left : right;  // 都在左子树或右子树
}
 
//10 求任意两结点距离
//首先找到两个结点的 LCA，然后分别计算 LCA 与它们的距离，最后相加即可。
 
int FindLevel(BTree node, BTree target)
{
    if (node == NULL)
        return -1;
    if (node == target)
        return 0;
  
    int level = FindLevel(node->lchild, target);  // 先在左子树找
    if (level == -1)
        level = FindLevel(node->rchild, target);  // 如果左子树没找到，在右子树找
  
    if (level != -1)  // 找到了，回溯
        return level + 1;
  
    return -1;  // 如果左右子树都没找到
}
 
int DistanceNodes(BTree node, BTree target1,BTree target2)
{
    BTree lca = FindLCA(node, target1, target2);  // 找到最低公共祖先结点
    int level1 = FindLevel(lca, target1); 
    int level2 = FindLevel(lca, target2);
  
    return level1 + level2;
}