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 thedefinition 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 allowa node to be a descendant of itself).”
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes2
and8
is6
. Another example is LCA of nodes2
and4
is2
, since a node can be a descendant of itself according to the LCA definition.
Thoughts:
Because of BST, we can decide which branch to to based on values. Two ways to approach this problem:
Iterative, O(1) space : Iteratively traversing down the side on which two nodes reside until the "split" is found.
Recuesive
Iterative:
Code 1
/**
* 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->val - p->val) * (root-> val - q->val) > 0)
root = (root-> val) > (p->val)? (root->left): (root-> right);
// =0 means either the current node is a. root is one of {p,q} b. root is the lowest parent of p and q.
return root;
}
};
Code 1 (Java)
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while ((root.val - p.val) * (root.val - q.val) > 0)
root = p.val < root.val ? root.left : root.right;
return root;
}
Code 1 (Python)
def lowestCommonAncestor(self, root, p, q):
while (root.val - p.val) * (root.val - q.val) > 0:
root = (root.left, root.right)[p.val > root.val]
return root
def lowestCommonAncestor(self, root, p, q):
while root:
if p.val < root.val > q.val:
root = root.left
elif p.val > root.val < q.val:
root = root.right
else:
return root
Recursive
Code 2
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
return (root-> val - p-> val)*(root ->val - q->val) > 0 ?
lowestCommonAncestor(p->val < root->val ? root->left: root->right, p , q): root;
}
};
Code 2 (Java)
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
return (root.val - p.val) * (root.val - q.val) > 0 ?
lowestCommonAncestor(p.val - root.val < 0 ? root.left: root.right, p , q) : root;
}
}
Code 2 (Python)
class Solution(object):
def lowestCommonAncestor(self, root, p, q):
return self.lowestCommonAncestor((root.left, root.right)[p.val > root.val], p, q) \
if (root.val - p.val) * (root.val - q.val) > 0 else root
Special thanks to StefanPochmann as he nailed this problem again over here.
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