原题链接 https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
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.
找到最近公共祖先
我使用两个hash表。
首先如果把root所在节点标记为1则，他的左儿子标记为2，右儿子标记为3，对于一个标记为x的节点，他的左儿子为x*2，右儿子为x*2+1;
所以我通过hash表PtoT，用标记映射地址。
另一个hash表VtoP，用值来映射标记。
当然可能存在相等的元素，因此我只记录最小的标记。
得到p和q的标记后，表可以将p和q中标记最大的除以2，直到p和q相等结束。
然后用Ptov[p]返回即可。
当然程序是可以优化的，可以在找到两个元素后便返回，不用记录所有点。
空间也可以进一步优化，只记录PtoT，而两个元素的标记在dfs的记录下来。
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
dfs(root, 1);
int pp = VtoP[p->val];
int pq = VtoP[q->val];
while (pp != pq)
{
if (pp > pq)pp >>= 1;
else pq >>= 1;
}
return PtoT[pp];
}
void dfs(TreeNode* root, int pos)
{
if (root == NULL)return;
if (PtoT.find(pos) == PtoT.end())
{
PtoT[pos] = root;
VtoP[root->val] = pos;
}
pos <<= 1;
dfs(root->left, pos);
dfs(root->right, pos + 1);
}
private:
unordered_map<int, TreeNode*>PtoT;
unordered_map<int, int>VtoP;
};