235 Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先
给定一棵二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
详见:https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/description/
Java实现:
方法一:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null){
return null;
}
if(root.val>Math.max(p.val,q.val)){
return lowestCommonAncestor(root.left,p,q);
}
if(root.val<Math.min(p.val,q.val)){
return lowestCommonAncestor(root.right,p,q);
}
return root;
}
}
方法二:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if(root==null||root==p||root==q){
return root;
}
TreeNode left=lowestCommonAncestor(root.left,p,q);
TreeNode right=lowestCommonAncestor(root.right,p,q);
if(left!=null&&right!=null){
return root;
}
return left!=null?left:right;
}
}
C++实现:
方法一:
/**
* 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==nullptr)
{
return root;
}
if(root->val>max(p->val,q->val))
{
return lowestCommonAncestor(root->left,p,q);
}
if(root->val<min(p->val,q->val))
{
return lowestCommonAncestor(root->right,p,q);
}
return root;
}
};
方法二:
/**
* 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==nullptr||root==p||root==q)
{
return root;
}
TreeNode *left=lowestCommonAncestor(root->left,p,q);
TreeNode *right=lowestCommonAncestor(root->right,p,q);
if(left&&right)
{
return root;
}
return left?left:right;
}
};
参考:http://www.cnblogs.com/grandyang/p/4640572.html
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