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.
分析：由于在所谓的二叉搜索树（binary search tree）中，处处满足顺序性（即任一节点的左（右）子树种，所有节点均小于（大于）r）。
思路： 1、当头结点为空时，返回空指针
2、如果节点p、q的值都比root的值要小，那么LCA一定为root的左子树；而如果节点p、q的值都比root的值要大，那么LCA一定为root的右子树；当节点p、q的值中一个比root的值要大，另一个比它小时，LCA就是root了。
3、我们要考虑是否能覆盖到节点是它自身子节点的情况，这时返回的是p或q的其中一个。
代码如下：（recursive solution）
/**
* 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) {
while(root !=nullptr){
if(p->val < root->val && q->val < root->val){
return lowestCommonAncestor(root->left,p,q);
}
else if(p->val > root->val && q->val > root->val){
return lowestCommonAncestor(root->right,p,q);
}
else return root; // 当p->val = root->val或 q->val =root->val时，就是节点是它自身子节点的情况了
}
return root;
}
};
也可以简洁点：
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (p -> val < root -> val && q -> val < root -> val)
return lowestCommonAncestor(root -> left, p, q);
if (p -> val > root -> val && q -> val > root -> val)
return lowestCommonAncestor(root -> right, p, q);
return root;
}
};
其他参考解法：
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if(p->val > root->val && q->val < root->val)
{
return root;
}
if(p->val < root->val && q->val > root->val)
{
return root;
}
if( p->val == root->val || q->val == root->val)
return root;
if( p->val < root->val && q->val < root->val)
return lowestCommonAncestor(root->left, p, q);
if( p->val > root->val && q->val > root->val)
return lowestCommonAncestor(root->right, p, q);
}
};
或：（iterative solution）
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
TreeNode* cur = root;
while (true) {
if (p -> val < cur -> val && q -> val < cur -> val)
cur = cur -> left;
else if (p -> val > cur -> val && q -> val > cur -> val)
cur = cur -> right;
else return cur;
}
}
};