235——Lowest Common Ancestor of a Binary Search Tree（二叉排序树）

Lowest Common Ancestor of a Binary Search Tree

Total Accepted: 7402 Total Submissions: 19069My Submissions
Question Solution

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.

Hide Tags
Tree

Hide Similar Problems
(M) Lowest Common Ancestor of a Binary Tree

Have you met this question in a real interview?
Yes

No

Discuss

思路，先判断入口是否有非法输入。

1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root

2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归

3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归

4 如果p<root<q，那么root肯定为LCA

以下代码仅为自己记录。

#include<iostream>
using namespace std;

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

TreeNode* L;
int digui(TreeNode* root,TreeNode* p,TreeNode* q)
{
if(root==p||root==q)
{
L=root;
return 0;
}
else
{
if(root->val<p->val&&root->val<q->val)
{
int a,b;
if(root->right!=NULL)
b=digui(root->right,p,q);
return a+b;
}
else if(root->val>p->val&&root->val>q->val)
{
int a,b;
if(root->left!=NULL)
b=digui(root->left,p,q);
return a+b;
}
else
{
L=root;
return 0;
}
}

}

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
digui(root,p,q);
return L;
}

int main()
{

}