235. Lowest Common Ancestor of a Binary Search Tree
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：
mine思路：
1.将各个节点的从root到节点的路径都记录下来vector<vector<int>>，vector中最后一个元素为当前节点。
2.找到目标的两个vector，比较之找到较短的"根"
others思路：
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。
代码如下：(代码实现为others思路)
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/**
* Definition for a binary tree node.
* 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) {
int
pVal = p->val;
int
qVal = q->val;
int
rootVal = root->val;
if
(pVal < rootVal && qVal < rootVal)
{
return
lowestCommonAncestor(root->left,p,q);
}
else
if
(pVal > rootVal && qVal > rootVal)
{
return
lowestCommonAncestor(root->right,p,q);
}
return
root;
}
};
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2016-08-07 09:47:25
