leetcode-235-Lowest Common Ancestor of a Binary Search Tree

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.

BST中，给出两个节点，求离它们最近的祖先。

因为是BST，root的左子树的值都小于root，root的右子树的值都大于root。

两个节点的公共路径的最后一个节点，即为最近的祖先。

递归

/**
* 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* solve(TreeNode* root,TreeNode* p,TreeNode* q){
if (p->val < root->val && q->val > root->val || p->val > root->val && q->val < root->val) return root;
if (p->val == root->val || q->val == root->val) return root;
if (p->val < root->val && q->val < root->val) return solve(root->left,p,q);
if (p->val > root->val && q->val > root->val) return solve(root->right,p,q);
}
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
return solve(root,p,q);
}
};

非递归

/**
* 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) {
while (root) {
if (p->val < root->val && q->val < root->val) root = root->left; // 向左
else if(p->val > root->val && q->val > root->val) root = root->right; //向右
else return root;
}
}
};