LCA系列
Problem I: Lowest Common Ancestor of a Binary Search Tree
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。
程序
public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == null) { return null; } if (root.val == p.val || root.val == q.val || (root.val > p.val && root.val < q.val) || (root.val < p.val && root.val > q.val)) { return root; } TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); return left == null ? right : left; } }
Problem II: Lowest Common Ancestor of a Binary Tree
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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).”
_______3______ / \ ___5__ ___1__ / \ / \ 6 _2 0 8 / \ 7 4
For example, the lowest common ancestor (LCA) of nodes 5
and 1
is 3
. Another example is LCA of nodes 5
and 4
is 5
, since a node can be a descendant of itself according to the LCA definition.
解决思路
递归,遍历的节点为输入的两个节点的其中一个则输出该节点。
程序
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 (root == p || root == q) { return root; } TreeNode left = lowestCommonAncestor(root.left, p, q); TreeNode right = lowestCommonAncestor(root.right, p, q); if (left != null && right != null) { return root; } return left == null ? right : left; } }