- 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]
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 nodes 2 and 8 is 6.
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 nodes 2 and 4 is 2, 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.
给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”
例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]
示例 1:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6
解释: 节点 2 和节点 8 的最近公共祖先是 6。
示例 2:
输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
说明:
所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉搜索树中。
[思路]：
此题的关键在于是二叉搜索树，即任一节点r的左（右）子树中，所有节点（若存在）均不大于（不小于）r。
所以p,q 比root小, 则LCA必定在左子树, 如果p,q比root大, 则LCA必定在右子树. 如果一大一小, 则root即为LCA. 利用迭代法。
[代码C++]：
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p,
TreeNode* q) {
if( root == NULL || p == NULL || q == NULL) return NULL;
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;
}
};