151 - 236 二叉树的最近公共祖先

题目

给定一个二叉树, 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，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]

img

示例 1:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 输出: 3 解释: 节点 5 和节点 1 的最近公共祖先是节点 3。

示例 2:

输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4 输出: 5 解释: 节点 5 和节点 4 的最近公共祖先是节点 5。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树中。

解答

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/discuss/65225/4-lines-C%2B%2BJavaPythonRuby

这个人巨骚的解法，代码就是要理解半天的那种。。

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

Runtime: 648 ms, faster than 5.01% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 37.6 MB, less than 5.55% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

虽然速度极慢。。

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: 76 ms, faster than 88.50% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 22.8 MB, less than 91.67% of Python3 online submissions for Lowest Common Ancestor of a Binary Tree.

正常写，速度就快了不少，而且可读性也好了不少。。

看来不能乱炫技

估计是在数组里面拿出来拿进去，很浪费内存和性能。

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: 64 ms, faster than 87.47% of JavaScript online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 41.4 MB, less than 75.00% of JavaScript 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 85.53% of Go online submissions for Lowest Common Ancestor of a Binary Tree.

Memory Usage: 7.5 MB, less than 100.00% of Go online submissions for Lowest Common Ancestor of a Binary Tree.