Lowest Common Ancestor of a Binary Search Tree(树中两个结点的最低公共祖先)

题目描述：

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

　　利用二叉排序树的性质，可以很好地解决这道题。

solution：

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (root == NULL || p == NULL || q == NULL)
        return NULL;

    if (root->val > p->val && root->val > q->val)
        return lowestCommonAncestor(root->left, p, q);
    else if (root->val < p->val && root->val < q->val)
        return lowestCommonAncestor(root->right, p, q);
    else 
        return root;
}

 　　如果只是一颗普通的二叉树呢？

solution：

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (!root)
        return NULL;
    if (root == p || root == q)
        return root;
    TreeNode *L = lowestCommonAncestor(root->left, p, q);
    TreeNode *R = lowestCommonAncestor(root->right, p, q);
    if (L && R)
        return root;
    return L ? L : R;  
}

来源：Lowest Common Ancestor of a Binary Tree Part I

    如果二叉树的结点存在指向父结点的指针，问题可以转化为求两个单链表的相交结点。

solution：

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode *parent;
};

int getHeight(TreeNode *p)
{
    int height = 0;
    while (p)
    {
        height++;
        p = p->parent;
    }
    return height;
}

TreeNode* lowestCommonAncestor(TreeNode *p, TreeNode *q)
{
    int h1 = getHeight(p);
    int h2 = getHeight(q);
    
    if (h1 > h2)
    {
        swap(h1, h2);
        swap(p, q);
    }

    int dh = h2 - h1;
    for (int h = 0; h < dh; ++h)
        q = q->parent;
    while (p && q)
    {
        if (p == q)
            return p;
        p = p->parent;
        q = q->parent;
    }
    return NULL;
}

来源：Lowest Common Ancestor of a Binary Tree Part II