Lowest Common Ancestor of a Binary Tree

LeetCode 236 | Difficulty: Medium

Medium

Problem Description

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, 10^5]`.

- `-10^9 <= Node.val <= 10^9`

- All `Node.val` are **unique**.

- `p != q`

- `p` and `q` will exist in the tree.

Topics: Tree, Depth-First Search, Binary Tree

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Approach

Tree DFS

Traverse the tree recursively (or with a stack). At each node, decide: what information do I need from the left/right subtrees? Process: go left → go right → combine results. Consider preorder, inorder, or postorder traversal based on when you need to process the node.

When to use

Path problems, subtree properties, tree structure manipulation.

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Solutions

Solution 1: C# (Best: 116 ms)

Metric	Value
Runtime	116 ms
Memory	N/A
Date	2018-04-25

Solution

/**
 * 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 == null || root==p || root==q) return root;
            var leftLca = LowestCommonAncestor(root.left, p, q);
            var rightLca = LowestCommonAncestor(root.right, p, q);
            if(leftLca==null) return rightLca;
            if(rightLca == null) return leftLca;
            return root;
    }
}

📜 1 more C# submission(s)

Submission (2018-07-27) — 152 ms, N/A

/**
 * 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 == null || root==p || root==q) return root;
            var leftLca = LowestCommonAncestor(root.left, p, q);
            var rightLca = LowestCommonAncestor(root.right, p, q);
            if(leftLca==null) return rightLca;
            if(rightLca == null) return leftLca;
            return root;
    }
}

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Complexity Analysis

Approach	Time	Space
Tree Traversal	$O(n)$	$O(h)$

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Interview Tips

Key Points

• Discuss the brute force approach first, then optimize. Explain your thought process.
• Consider: "What information do I need from each subtree?" — this defines your recursive return value.