README_EN.md

235. Lowest Common Ancestor of a Binary Search Tree

中文文档

Description

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given binary search tree:  root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8

Output: 6

Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4

Output: 2

Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 

Note:

• All of the nodes' values will be unique.
• p and q are different and both values will exist in the BST.

Solutions

Python3

Iterative:

# 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 root:
            if root.val < p.val and root.val < q.val:
                root = root.right
            elif root.val > p.val and root.val > q.val:
                root = root.left
            else:
                return root

Recursive:

# 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 is None:
            return None
        if root.val < p.val and root.val < q.val:
            return self.lowestCommonAncestor(root.right, p, q)
        if root.val > p.val and root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        return root

Java

Iterative:

/**
 * 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 (root != null) {
            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;
        }
        return root;
    }
}

Recursive:

/**
 * 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 == null) return null;
        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;
    }
}

Go

Iterative:

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

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	for root != nil {
		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
		}
	}
	return nil
}

Recursive:

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

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil {
        return root
    }
    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
}

...