leetcode - Lowest Common Ancestor of a Binary Search Tree

leetcode - 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.

/**
* 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 small, big;
if(p->val < q->val){
small = p->val;
big = q->val;
}
else{
small = q->val;
big = p->val;
}
// small = min(p->val, q->val);
// big = max(p->val, q->val);
if(root->val >= small && root->val <= big) return root;
else if(root->val < small) return lowestCommonAncestor(root->right, p, q);
else return lowestCommonAncestor(root->left, p, q);
}
};

题意为找到两个节点最低的公共祖先节点。经过分析可以发现，如果一个节点左右子树中分别包含两个节点，那么这个节点一定是LCA(lowest common ancestor)。

由于是二叉搜索树，有性质：左子树节点小于根节点，右子树节点大于根节点。所以可以比较当前节点与两个节点的大小判断，从根开始遍历，第一个满足大小位于两数之间的节点，必定其“左右子树中分别包含两个节点”。