72 - 二叉树的最近公共祖先

题目

给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]

img

示例 1:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 输出: 6 解释: 节点 2 和节点 8 的最近公共祖先是 6。

示例 2:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 输出: 2 解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。

说明:

• 所有节点的值都是唯一的。

• p、q 为不同节点且均存在于给定的二叉搜索树中。

解答

递归

因为二叉树是根据大小排列的，因此可以根据p与q的值来分类递归求解

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root == None or root == p or root == q:
            return root
        left = Solution.lowestCommonAncestor(self, root.left, p, q)
        right = Solution.lowestCommonAncestor(self, root.right, p, q)
        if left and right:
            return root
        return left if left else right

Runtime: 68 ms, faster than 87.92% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 23.3 MB, less than 91.67% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    left := lowestCommonAncestor(root.Left, p, q)
    right := lowestCommonAncestor(root.Right, p, q)
    if left != nil && right != nil {
        return root
    }
    if left != nil {
        return left
    } else {
        return right
    }
}

Runtime: 12 ms, faster than 85.78% of Go online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 7.6 MB, less than 100.00% of Go online submissions for Lowest Common Ancestor of a Binary Tree.

var lowestCommonAncestor = function(root, p, q) {
  if (!root || root === p || root === q) {
    return root
  }
  const left = lowestCommonAncestor(root.left, p, q)
  const right = lowestCommonAncestor(root.right, p, q)
  if (left && right) {
    return root
  }
  return left ? left : right
};

Runtime: 68 ms, faster than 78.64% of JavaScript online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 41.6 MB, less than 50.00% of JavaScript online submissions for Lowest Common Ancestor of a Binary Tree.

光头大佬的四行解法

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        if root in (None, p, q): return root
        subs = [self.lowestCommonAncestor(kid, p, q)
            for kid in (root.left, root.right)]
        return root if all(subs) else max(subs)

Runtime: 80 ms, faster than 15.07% of Python online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 26.9 MB, less than 5.88% of Python online submissions for Lowest Common Ancestor of a Binary Tree.