leetcode 235: Lowest Common Ancestor of a Binary Search Tree

最新推荐文章于 2025-06-16 22:55:50 发布

mikefan1991

最新推荐文章于 2025-06-16 22:55:50 发布

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本文介绍了一种优化后的二叉树最近公共祖先查找算法，通过提前剪枝减少不必要的递归调用，提高了搜索效率。适用于解决在二叉树中查找两个节点的最近公共祖先的问题。

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Use recursion. if none of root, root->left and root->right equal to p or q, return NULL. If left and right are both not NULL, this root is the ancestor. If root equals to p or q, return the root to represent that at least one node is found. If left or right is not NULL, return it to represent at least one node is found in the subtree of this root.

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

The code above can be accepted in the Lowest Common Ancestor of a Binary Tree. But it is not the fastest way in this problem. So I add a few conditions to do the pruning. The following is the updated code:

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