[LeetCode] 235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先
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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes2and8is6.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes2and4is2, since a node can be a descendant of itself according to the LCA definition.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the BST.
求二叉搜索树(BST)的最近公共祖先(LCA)。最近公共祖先是指在一个树或者有向无环图中同时拥有v和w作为后代的最深的节点。
解法1：递归，
1. P, Q都比root小，则LCA在左树，我们继续在左树中寻找LCA
2. P, Q都比root大，则LCA在右树，我们继续在右树中寻找LCA
3. 其它情况，表示P,Q在root两边，或者二者其一是root，或者都是root，这些情况表示root就是LCA，直接返回root即可。
解法2: 迭代
判断标准同解法1，只是用迭代来实现。
Java:
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
if(root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
return root;
}
}
Java:
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
//发现目标节点则通过返回值标记该子树发现了某个目标结点
if(root == null || root == p || root == q) return root;
//查看左子树中是否有目标结点，没有为null
TreeNode left = lowestCommonAncestor(root.left, p, q);
//查看右子树是否有目标节点，没有为null
TreeNode right = lowestCommonAncestor(root.right, p, q);
//都不为空，说明做右子树都有目标结点，则公共祖先就是本身
if(left!=null&&right!=null) return root;
//如果发现了目标节点，则继续向上标记为该目标节点
return left == null ? right : left;
}
}
Python:
class Solution:
# @param {TreeNode} root
# @param {TreeNode} p
# @param {TreeNode} q
# @return {TreeNode}
def lowestCommonAncestor(self, root, p, q):
s, b = sorted([p.val, q.val])
while not s <= root.val <= b:
# Keep searching since root is outside of [s, b].
root = root.left if s <= root.val else root.right
# s <= root.val <= b.
return root
C++:
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root) return NULL;
if (root->val > max(p->val, q->val))
return lowestCommonAncestor(root->left, p, q);
else if (root->val < min(p->val, q->val))
return lowestCommonAncestor(root->right, p, q);
else return root;
}
};
C++:
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
while (true) {
if (root->val > max(p->val, q->val)) root = root->left;
else if (root->val < min(p->val, q->val)) root = root->right;
else break;
}
return root;
}
};
类似题目：
[LeetCode] 236. Lowest Common Ancestor of a Binary Tree 二叉树的最近公共祖先
