leetcode 235. Lowest Common Ancestor of a Binary Search Tree 最近公共祖先 + BST
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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 v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______6______ / \___2__ ___8__/ \ / \0 _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.
这道题的主要问题是搞懂LCA最近公共祖先的定义。
LCA最近公共祖先,指的是给定一个树tree和查询接单a和b,LCA就是找到距离root结点最远的并且是a和b的祖先结点
在二叉查找树种,寻找两个节点的最低公共祖先。
1、如果a、b都比根节点小,则在左子树中递归查找公共节点。
2、如果a、b都比根节点大,则在右子树中查找公共祖先节点。
3、如果a、b一个比根节点大,一个比根节点小,或者有一个等于根节点,则根节点即为最低公共祖先。
建议和这道题一起学习leetcode 236. Lowest Common Ancestor of a Binary Tree 最近公告祖先LCA + 二叉树 一起学习。
代码如下:
/*class TreeNode{ int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; }}*//* * LCA最近公共祖先,指的是给定一个树tree和查询接单a和b,LCA就是找到 * 距离root结点最远的并且是a和b的祖先结点 * * 在二叉查找树种,寻找两个节点的最低公共祖先。 * 1、如果a、b都比根节点小,则在左子树中递归查找公共节点。 * 2、如果a、b都比根节点大,则在右子树中查找公共祖先节点。 * 3、如果a、b一个比根节点大,一个比根节点小,或者有一个等于根节点,则根节点即为最低公共祖先。 * * */public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root==null || p==null || q==null) return null; else 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; }}
下面是C++的做法,就是对BFS二叉搜索树的一个简单的DFS深度优先搜索的遍历,要搞清楚LCA的含义
代码如下:
#include <iostream>#include <algorithm>#include <vector>#include <set>#include <string>#include <map>using namespace std;/*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 == NULL || p == NULL || q == NULL) return NULL; else 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; }};
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