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 allowa node to be a descendant of itself).”
_______6______
/ ___2__ ___8__
/ \ / 0 _4 7 9
/ 3 5
For example, the lowest common ancestor (LCA) of nodes2 and
8 is 6. Another example is LCA of nodes2 and
4 is 2, since a node can be a descendant of itself according to the LCA definition.
分析:
求二叉树节点最低公共祖先的题目,应该做到以下两点:
1、时间复杂度O(n),只需遍历一遍二叉树。
2、要判断输入的两个节点是否在二叉树中。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
struct node
{
TreeNode* res;
bool bp,bq;
node(TreeNode* n,bool x,bool y):res(n),bp(x),bq(y){}
};
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q)
{
if(!root || !p || !q)
return nullptr;
return lowestCommonAncestor_core(root,p,q).res;
}
node lowestCommonAncestor_core(TreeNode* root, TreeNode* p, TreeNode* q)
{
if(root==nullptr)
return node(nullptr,false,false);
if(root==p)
{
return hasnode(root,q)?node(root,true,true):node(root,true,false);
}
if(root==q)
{
return hasnode(root,p)?node(root,true,true):node(root,false,true);
}
if(root->left)
{
auto r=lowestCommonAncestor_core(root->left,p,q);
if(r.bp && r.bq)
return r;
else if(!r.bp && r.bq)
{
if(hasnode(root->right,p))
return node(root,true,true);
else
return node(nullptr,false,true);
}
else if(r.bp && !r.bq)
{
if(hasnode(root->right,q))
return node(root,true,true);
else
return node(nullptr,true,false);
}
}
return lowestCommonAncestor_core(root->right,p,q);
}
bool hasnode(TreeNode* root, TreeNode* p)
{
if(root==p)
return true;
if(root==nullptr)
return false;
return hasnode(root->left,p) || hasnode(root->right,p);
}
};版权声明:本文为博主原创文章,未经博主允许不得转载。
leetcode - Lowest Common Ancestor of a Binary Search Tree
原文:http://blog.csdn.net/bupt8846/article/details/46840209