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

jmesSehng
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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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