LeetCode - 235 Lowest Common Ancestor of a Binary Search Tree

作者 QIFAN 日期 2016-07-13
LeetCode - 235 Lowest Common Ancestor of a Binary Search Tree

Difficulty:Easy

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

首先要理解BST的概念,即对于某一节点root的值,肯定大于其左侧树所有节点的值,而小于右侧数所有节点的值。这样就可以通过比较val来找到目标。

一开始没有get到BST的精髓,起码被坑了3个小时。

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) return null;
if(p==null) return q;
if(q==null) return p;
if(p.val==q.val) return q;
TreeNode left = p.val<q.val?p:q;
TreeNode right = left.equals(p) ? q:p;
return search(root, left, right);
}
public TreeNode search(TreeNode root, TreeNode left, TreeNode right) {
if(root.val==left.val||root.val==right.val) return root;
if(right.val < root.val) return search(root.left, left, right);
if(left.val > root.val) return search(root.right, left, right);
return root;
// return (root.val==left.val || root.val==right.val)?root:(right.val<root.val?search(root.left, left, right):(left.val>root.val?search(root.right,left, right):root));
}
}

原题链接:https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/