235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

 

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [2,1], p = 2, q = 1
Output: 2

 

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

A:
根据val ， 看左边右边

public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(p.val > q.val)
            return lowestCommonAncestor(root,q,p);
        return helper(root,p,q);
    }
    private TreeNode helper(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val<p.val)
            return helper(root.right,p,q);
        else if(root.val > q.val)
            return helper(root.left,p,q);
        return root;
    }
}

其实这道题，无关Binary Search Tree ---虽然利用起来，效率会高一些些

/**
 * 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)
            return root;
        if(root->val == p->val || root->val == q->val){
            return root;
        }
        auto l = lowestCommonAncestor(root->left, p, q);
        auto r = lowestCommonAncestor(root->right, p, q);
        if(l&&r){
            return root;
        }else{
            return l?l:r;
        }
    }
};

Mistakes: