[LeetCode]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.

用到二叉搜索树本问题就很好解。

二叉搜索树的特点就是root->leff < root < root->right。

如果root>p && root>q，那么LCA在左子树上；

如果root<p && root<q，那么LCA在右子树上。

递归实现。

/**
 * 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 || !p || !q) return NULL;
        if(root->val>p->val && root->val>q->val) return lowestCommonAncestor(root->left,p,q);
        if(root->val<p->val && root->val<q->val) return lowestCommonAncestor(root->right,p,q);
        return root;
    }
};

迭代实现。

/**
 * 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 || !p || !q) return NULL;
        TreeNode *cur = root;
        while(cur)
        {
            if(cur->val>p->val && cur->val>q->val) cur = cur->left;
            else if(cur->val<p->val && cur->val<q->val) cur = cur->right;
            else return cur;
        }
        return cur;
    }
};