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1. Title
235. Lowest Common Ancestor of a Binary Search Tree
2. Http address
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/
3. The question
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.
4 My code(AC)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean DFS(TreeNode root, TreeNode target, LinkedList<TreeNode> path) { if (root == null || target == null) return false; path.addLast(root); if (root == target) { return true; } if (DFS(root.left, target, path)) { return true; } else { if (root.left != null) { path.removeLast(); } boolean re = DFS(root.right, target, path); if( root.right != null && re == false){ path.removeLast(); } return re; } } public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if( p == null || q == null) { return null; } LinkedList<TreeNode> path1 = new LinkedList<TreeNode>(); LinkedList<TreeNode> path2 = new LinkedList<TreeNode>(); DFS(root,p,path1); DFS(root,q,path2); if( path1.size() <= 0 || path2.size() <= 0) return null; TreeNode bf = path1.removeFirst(); path2.removeFirst(); while( !path1.isEmpty() && !path2.isEmpty()){ TreeNode node1 = path1.removeFirst(); TreeNode node2 = path2.removeFirst(); if( node1 != node2){ break; } bf = node1; } return bf; } }
Lowest Common Ancestor of a Binary Search Tree
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原文地址：http://www.cnblogs.com/ordili/p/5515744.html
