LeetCode Algorithms #235 <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.

思路：

找到一点，它的值正好在要找的两子节点的值的中间。

如果发现当前选中的点，比两个子节点的值都大，那么要找的点在当前点的左子树中。

如果发现当前选中的点，比两个子节点的值都笑，那么要找的点在当前点的右子树中。

解：

/**
* 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 == NULL || p == NULL || q == NULL)
return NULL;

if(p == root)
return p;
else if(q == root)
return q;

TreeNode* lagerNode = NULL;
TreeNode* smallerNode = NULL;
if(p->val >= root->val)
{
lagerNode = p;
smallerNode = q;
}
else
{
lagerNode = q;
smallerNode = p;
}

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

}
};