原題鏈接 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
我使用兩個hash表。
首先如果把root所在節點標記爲1則，他的左兒子標記爲2，右兒子標記爲3，對於一個標記爲x的節點，他的左兒子爲x*2，右兒子爲x*2+1;
所以我通過hash表PtoT，用標記映射地址。
另一個hash表VtoP，用值來映射標記。
當然可能存在相等的元素，因此我只記錄最小的標記。
得到p和q的標記後，表可以將p和q中標記最大的除以2，直到p和q相等結束。
然後用Ptov[p]返回即可。
當然程序是可以優化的，可以在找到兩個元素後便返回，不用記錄所有點。
空間也可以進一步優化，只記錄PtoT，而兩個元素的標記在dfs的記錄下來。
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;
};