【LeetCode】235. Lowest Common Ancestor of a Binary Search Tree (2 solutions)

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.

解法一：不考虑BST的特性，对于一棵普通二叉树，寻找其中两个节点的LCA

深度遍历到节点p时，栈中的所有节点即为p的从根开始的祖先序列。

因此只需要比较p、q祖先序列中最后一个相同的祖先即可。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// special cases
if(root == NULL)
return NULL;
if(p == root || q == root)
return root;
if(p == q)
return p;

vector<TreeNode*> vp;
vector<TreeNode*> vq;
stack<TreeNode*> stk;
unordered_map<TreeNode*, bool> m;   //visited
stk.push(root);
m[root] = true;
while(!stk.empty())
{
TreeNode* top = stk.top();
if(top->left && m[top->left] == false)
{
stk.push(top->left);
m[top->left] = true;
if(top->left == p)
{
vp = stkTovec(stk);
if(!vq.empty())
break;
}
if(top->left == q)
{
vq = stkTovec(stk);
if(!vp.empty())
break;
}
continue;
}
if(top->right && m[top->right] == false)
{
stk.push(top->right);
m[top->right] = true;
if(top->right == p)
{
vp = stkTovec(stk);
if(!vq.empty())
break;
}
if(top->right == q)
{
vq = stkTovec(stk);
if(!vp.empty())
break;
}
continue;
}
stk.pop();
}
int i = 0;
for(; i < vp.size() && i < vq.size(); i ++)
{
if(vp[i] != vq[i])
break;
}
return vp[i-1];
}
vector<TreeNode*> stkTovec(stack<TreeNode*> stk)
{
vector<TreeNode*> v;
while(!stk.empty())
{
TreeNode* top = stk.top();
stk.pop();
v.push_back(top);
}
reverse(v.begin(), v.end());
return v;
}
};

解法二：利用BST的性质，

p和q的LCA是恰好把p、q分叉的节点，也就是LCA的值介于p、q之间

从最高层的祖先root开始往下，

若不满足LCA条件，往p、q所在子树递归。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if((root->val - p->val) * (root->val - q->val) <= 0)
return root;
else if(root->val > p->val)
return lowestCommonAncestor(root->left, p, q);
else
return lowestCommonAncestor(root->right, p, q);
}
};