LCA of BST
描述
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
_______6______
/ \
___2__ ___8__
/ \ / \
1 _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
根据二叉搜索树的性质，两个子节点p,q和根节点root的关系，有以下四种情况：
- 两个子节点都在树的左子树上
- 两个子节点都在树的右子树上
- 一个子节点在左子树，一个子节点在右子树
- 一个子节点的值和根节点的值相等
以题目中的树为例，节点1和节点4为情况1，节点7和节点9为情况2，节点1和节点7为情况3，节点2和4为情况4。 若为情况3或4，当前节点即为最近公共祖先，若为情况1或2，则还需递归到左或右子树上，继续这个过程。
该算法的时间复杂度为O(h)，h为树的高度。
解法1 递归
// LCA of BST
// Time Complexity: O(h), Space Complexity: O(h)
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) return null;
if (Math.max(p.val, q.val) < root.val) {
return lowestCommonAncestor(root.left, p, q);
} else if (Math.min(p.val, q.val) > root.val) {
return lowestCommonAncestor(root.right, p, q);
} else {
return root;
}
}
}解法2 迭代
// LCA of BST
// Time Complexity: O(h), Space Complexity: O(1)
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
while (root != null) {
if (Math.max(p.val, q.val) < root.val) {
root = root.left;
} else if (Math.min(p.val, q.val) > root.val) {
root = root.right;
} else {
return root;
}
}
return null;
}
}