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leetcode235 - Lowest Common Ancestor of a Binary Search Tree

发布时间:2020-12-14 04:21:14 所属栏目:大数据 来源:网络整理
导读: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 p and q as the lowest node in

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 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).”
Given binary search tree: root = [6,2,8,4,7,9,null,3,5]
_______6______
/
___2__ ___8__
/ /
0 _4 7 9
/
3 5

Example 1:
Input: root = [6,5],p = 2,q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
Example 2:
Input: root = [6,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.
Note:
* All of the nodes‘ values will be unique.
* p and q are different and both values will exist in the BST.

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利用BST特性。
1.如果root值在要找的两个数中间,说明两个节点分布在左右子树里,当前的根就是LCA。
2.如果root值和两个数其中一个相等,说明两个数的节点是亲子关系,当前的根就是LCA。
以上面两条规则为递归出口,然后做递归。
类似问题:Lowest Common Ancestor of a Binar Tree:?https://www.cnblogs.com/jasminemzy/p/9698483.html

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实现:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q) {
        if (root == null || root.val == p.val || root.val == q.val) {
            return root;
        }
        if (root.val > p.val && root.val < q.val || root.val < p.val && root.val > q.val) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left,p,q);
        TreeNode right = lowestCommonAncestor(root.right,q);
        return left != null ? left : right;
    }
}

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