235. 二叉搜索树的最近公共祖先 

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

我们从根节点开始遍历，如果当前节点的值小于 $\textit{p}$ 和 $\textit{q}$ 的值，说明 $\textit{p}$ 和 $\textit{q}$ 应该在当前节点的右子树，因此将当前节点移动到右子节点；如果当前节点的值大于 $\textit{p}$ 和 $\textit{q}$ 的值，说明 $\textit{p}$ 和 $\textit{q}$ 应该在当前节点的左子树，因此将当前节点移动到左子节点；否则说明当前节点就是 $\textit{p}$ 和 $\textit{q}$ 的最近公共祖先，返回当前节点即可。

时间复杂度 $O(n)$，其中 $n$ 是二叉搜索树的节点个数。空间复杂度 $O(1)$。

Python3

# 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

Java

/**
 * 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;
            }
        }
    }
}

C++

/**
 * 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;
            }
        }
    }
};

Go

/**
 * 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 < min(p.Val, q.Val) {
			root = root.Right
		} else if root.Val > max(p.Val, q.Val) {
			root = root.Left
		} else {
			return root
		}
	}
}

TypeScript

/**
 * 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;
        }
    }
}

C#

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

public 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;
            }
        }
    }
}

方法二：递归

我们也可以使用递归的方法来解决这个问题。

我们首先判断当前节点的值是否小于 $\textit{p}$ 和 $\textit{q}$ 的值，如果是，则递归遍历右子树；如果当前节点的值大于 $\textit{p}$ 和 $\textit{q}$ 的值，如果是，则递归遍历左子树；否则说明当前节点就是 $\textit{p}$ 和 $\textit{q}$ 的最近公共祖先，返回当前节点即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉搜索树的节点个数。

Python3

# 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

Java

/**
 * 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;
    }
}

C++

/**
 * 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;
    }
};

Go

/**
 * 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
}

TypeScript

/**
 * 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;
}

C#

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

public 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;
    }
}