题目

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.

思路

[算法系列之三十一]近期公共祖先（LCA）

代码

/*---------------------------------------
*   日期：2015-07-14
*   作者：SJF0115
*   题目: 235.Lowest Common Ancestor of a Binary Search Tree
*   网址：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
*   结果：AC
*   来源：LeetCode
*   博客：
-----------------------------------------*/
#include <iostream>
#include <vector>
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 == nullptr || p == nullptr || q == nullptr){return nullptr;}//ifreturn helper(root,p,q);}
private:TreeNode* helper(TreeNode* root,TreeNode* p,TreeNode* q){if(root == nullptr || p == nullptr || q == nullptr){return nullptr;}//ifint pVal = p->val;int qVal = q->val;int rootVal = root->val;// 分居两側if((pVal <= rootVal && qVal >= rootVal) || (pVal >= rootVal && qVal <= rootVal)){return root;}//if// 左側if(pVal < rootVal && qVal < rootVal){return helper(root->left,p,q);}//if// 右側if(pVal > rootVal && qVal > rootVal){return helper(root->right,p,q);}//if}
};int main(){Solution s;TreeNode* root = new TreeNode(6);TreeNode* node1 = new TreeNode(0);TreeNode* node2 = new TreeNode(9);TreeNode* node3 = new TreeNode(2);TreeNode* node4 = new TreeNode(3);TreeNode* node5 = new TreeNode(4);TreeNode* node6 = new TreeNode(5);TreeNode* node7 = new TreeNode(7);TreeNode* node8 = new TreeNode(8);root->left = node3;root->right = node8;node3->left = node1;node3->right = node5;node5->left = node4;node5->right = node6;node8->left = node7;node8->right = node2;TreeNode* node = s.lowestCommonAncestor(root,node3,node4);if(node != nullptr){cout<<node->val<<endl;}//ifelse{cout<<"nullptr"<<endl;}//elsereturn 0;
}

执行时间