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

2016-07-30 11:09 555 查看

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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