leetcode_235——Lowest Common Ancestor of a Binary Search Tree（二叉排序树）
Lowest Common Ancestor of a Binary Search Tree
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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.
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()
{
}