Question

235. 二叉搜索树的最近公共祖先

English Version

题目描述

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]

 

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。

解法

方法一：迭代或递归

从上到下搜索，找到第一个值位于 $[p.val, q.val]$ 之间的结点即可。

既可以用迭代实现，也可以用递归实现。

迭代的时间复杂度为 $O(n)$，空间复杂度为 $O(1)$。

递归的时间复杂度为 $O(n)$，空间复杂度为 $O(n)$。

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def lowestCommonAncestor(
    self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'
  ) -> 'TreeNode':
    while 1:
      if root.val < min(p.val, q.val):
        root = root.right
      elif root.val > max(p.val, q.val):
        root = root.left
      else:
        return root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */

class Solution {
  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    while (true) {
      if (root.val < Math.min(p.val, q.val)) {
        root = root.right;
      } else if (root.val > Math.max(p.val, q.val)) {
        root = root.left;
      } else {
        return root;
      }
    }
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
  TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    while (1) {
      if (root->val < min(p->val, q->val)) {
        root = root->right;
      } else if (root->val > max(p->val, q->val)) {
        root = root->left;
      } else {
        return root;
      }
    }
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val   int
 *   Left  *TreeNode
 *   Right *TreeNode
 * }
 */

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
  for {
    if root.Val < p.Val && root.Val < q.Val {
      root = root.Right
    } else if root.Val > p.Val && root.Val > q.Val {
      root = root.Left
    } else {
      return root
    }
  }
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function lowestCommonAncestor(
  root: TreeNode | null,
  p: TreeNode | null,
  q: TreeNode | null,
): TreeNode | null {
  while (root) {
    if (root.val > p.val && root.val > q.val) {
      root = root.left;
    } else if (root.val < p.val && root.val < q.val) {
      root = root.right;
    } else {
      return root;
    }
  }
}

方法二

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, x):
#     self.val = x
#     self.left = None
#     self.right = None


class Solution:
  def lowestCommonAncestor(
    self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'
  ) -> 'TreeNode':
    if root.val < min(p.val, q.val):
      return self.lowestCommonAncestor(root.right, p, q)
    if root.val > max(p.val, q.val):
      return self.lowestCommonAncestor(root.left, p, q)
    return root

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode(int x) { val = x; }
 * }
 */

class Solution {
  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root.val < Math.min(p.val, q.val)) {
      return lowestCommonAncestor(root.right, p, q);
    }
    if (root.val > Math.max(p.val, q.val)) {
      return lowestCommonAncestor(root.left, p, q);
    }
    return root;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
  TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
    if (root->val < min(p->val, q->val)) {
      return lowestCommonAncestor(root->right, p, q);
    }
    if (root->val > max(p->val, q->val)) {
      return lowestCommonAncestor(root->left, p, q);
    }
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val   int
 *   Left  *TreeNode
 *   Right *TreeNode
 * }
 */

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
  if root.Val < p.Val && root.Val < q.Val {
    return lowestCommonAncestor(root.Right, p, q)
  }
  if root.Val > p.Val && root.Val > q.Val {
    return lowestCommonAncestor(root.Left, p, q)
  }
  return root
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function lowestCommonAncestor(
  root: TreeNode | null,
  p: TreeNode | null,
  q: TreeNode | null,
): TreeNode | null {
  if (root.val > p.val && root.val > q.val) return lowestCommonAncestor(root.left, p, q);
  if (root.val < p.val && root.val < q.val) return lowestCommonAncestor(root.right, p, q);
  return root;
}