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  • [LeetCode] 235. 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 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.

    二叉搜索树的最小公共祖先。

    影子题236。题意是给一个二叉搜索树(BST)和两个节点p和q,请找出两个节点的最小公共祖先。思路是递归,因为BST的性质,所以任何节点的左孩子一定比当前节点小,右孩子一定比当前节点大。明白这一点,代码就不难写。

    时间O(n)

    空间O(n)

    Java实现

     1 class Solution {
     2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
     3         if (root.val > p.val && root.val > q.val) {
     4             return lowestCommonAncestor(root.left, p, q);
     5         } else if (root.val < p.val && root.val < q.val) {
     6             return lowestCommonAncestor(root.right, p, q);
     7         } else {
     8             return root;
     9         }
    10     }
    11 }

    JavaScript实现

     1 /**
     2  * @param {TreeNode} root
     3  * @param {TreeNode} p
     4  * @param {TreeNode} q
     5  * @return {TreeNode}
     6  */
     7 var lowestCommonAncestor = function (root, p, q) {
     8     if (root.val > p.val && root.val > q.val) {
     9         return lowestCommonAncestor(root.left, p, q);
    10     } else if (root.val < p.val && root.val < q.val) {
    11         return lowestCommonAncestor(root.right, p, q);
    12     } else {
    13         return root;
    14     }
    15 };

    相关题目

    235. Lowest Common Ancestor of a Binary Search Tree

    236. Lowest Common Ancestor of a Binary Tree

    865. Smallest Subtree with all the Deepest Nodes

    1257. Smallest Common Region

    LeetCode 题目总结

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  • 原文地址:https://www.cnblogs.com/cnoodle/p/12455547.html
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