LeetCode-in-All

236. Lowest Common Ancestor of a Binary Tree

Medium

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

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).”

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1

Output: 3

Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4

Output: 5

Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

Input: root = [1,2], p = 1, q = 2

Output: 1

Constraints:

• The number of nodes in the tree is in the range [2, 105].
• -109 <= Node.val <= 109
• All Node.val are unique.
• p != q
• p and q will exist in the tree.

Solution

import com_github_leetcode.TreeNode

/*
 * Definition for a binary tree node.
 * class TreeNode(var _value: Int) {
 *   var value: Int = _value
 *   var left: TreeNode = null
 *   var right: TreeNode = null
 * }
 */
object Solution {
    def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
        if (root == null) {
            return null
        }
        if (root.value == p.value || root.value == q.value) {
            return root
        }
        val left = lowestCommonAncestor(root.left, p, q)
        val right = lowestCommonAncestor(root.right, p, q)
        if (left != null && right != null) {
            return root
        }
        if (left != null) {
            return left
        }
        right
    }
}

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