问题描述
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.
解决思路
递归,可以用BST特有的二分递归,也可以普通递归。
程序
普通:
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return root; } if (p == null) { return q; } if (q == null) { return p; } if (p == q) { return p; } return helper(root, p, q); } private static TreeNode helper(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return null; } if (root == p || root == q) { return root; } TreeNode left = helper(root.left, p, q); TreeNode right = helper(root.right, p, q); if (left != null && right != null) { return root; } return left == null ? right : left; } }
BST特性:
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return root; } if (p == null) { return q; } if (q == null) { return p; } if (p == q) { return p; } return helper(root, p, q); } private static TreeNode helper(TreeNode root, TreeNode p, TreeNode q) { if (p.val > root.val && q.val > root.val) { return helper(root.right, p, q); } if (p.val < root.val && q.val < root.val) { return helper(root.left, p, q); } return root; } }