leetcode刷题日记——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.

问题分析：题目所要达到的目标就是在二叉查找树中找到给定二个节点的最近祖先。由于是二叉查找树，不妨先分析所要找的祖先节点满足的性质，可以推出它对应的值是介于二个节点的值得中间， p（q）<=root<=q(p),做此题要避免一个误区，不要想当然的认为p的值小于q,所以介于关系要看下面代码中的实现。如果二个节点位于当前判定节点的同一侧，则对该侧进行递归，不过我这里解法还没考虑最优解。代码如下：

/**
* 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) {
if((root->val-p->val)*(root->val-q->val)<=0)
return root;
else if(p->val<root->val&&q->val<root->val)
return lowestCommonAncestor( root->left,  p,  q);
else
return lowestCommonAncestor( root->right,  p,  q);

}
};