【LeetCode】235 Lowest Common Ancestor of a Binary Search Tree

寻找二叉搜索树中两个节点的最近公共祖先

最新推荐文章于 2024-10-16 01:06:12 发布

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最新推荐文章于 2024-10-16 01:06:12 发布

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分类专栏：算法 LeetCode 二叉树 LeetCode解题报告

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本文详细介绍了如何在一个给定的二叉搜索树中找到两个节点的最近公共祖先，并通过Java代码实现了解决方案。

Lowest Common Ancestor of a Binary Search Tree
Total Accepted: 3808 Total Submissions: 9820 My Submissions Question Solution
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.

【解题思路】
二叉搜索树，根据特点搜索，比当前值大，搜索右子树，比当前值小，搜索左子树，相等就返回结果。

Java AC

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        List<TreeNode> list1 = new ArrayList<TreeNode>();
        List<TreeNode> list2 = new ArrayList<TreeNode>();
        getPath(root, p, list1);
        getPath(root, q, list2);
        int size1 = list1.size();
        int size2 = list2.size();
        int len = size1 < size2 ? size1 : size2;
        int k = 0;
        while (k < len){
            TreeNode node1 = list1.get(k);
            TreeNode node2 = list2.get(k);
            if (node1 != node2){
                break;
            }
            k++;
        }
        return list1.get(k - 1);
    }
    
    private void getPath(TreeNode root, TreeNode d, List<TreeNode> list){
        if (root != null){
            int rval = root.val;
            int dval = d.val;
            list.add(root);
            if (dval < rval){
                getPath(root.left, d, list);
            }else if (dval > rval){
                getPath(root.right, d, list);
            }
        }
    }
}