72【leetcode】经典算法- Lowest Common Ancestor of a Binary Search Tree（lct of bst）

待我称王封你为后i 2022-09-24 11:29 135阅读 0赞

# 题目描述： #

## 一个二叉搜索树，给定两个节点a，b，求最小的公共祖先 ##

_______6______
           /              \
        ___2__          ___8__
       /      \        /      \
       0      _4       7       9
             /  \
             3   5

## 例如： ##

2,8 ---->6 2,4----->2

# 原文描述： #

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.

思路分析：

*  二叉树考虑递归的思路
 *  如果a < = root <= b，可以确定root是lct
 *  采用二分搜索的方式递归，root > a,b，继续遍历左子树，反之右边的子树  
    \-在递归的开始判断空值的情况，直接返回root

# 代码： #

/**
     * 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) {
            if(root == null || p == null || q == null)
                return root;
            if(p.val > root.val && q.val > root.val)
                return lowestCommonAncestor(root.right, p, q);
            else if(p.val < root.val && q.val < root.val)
                return lowestCommonAncestor(root.left, p, q);
            else
                return root;
        }
    }

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[leetcode]: http://blog.csdn.net/column/details/fengsheng.html