LeetCode 236. Lowest Common Ancestor of a Binary Tree

Question

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.

Source: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/

Solution

DFS. Use the return value of recursive function to reflect the existence of p, q and their lowest common ancestor. The following implementation is feasible but no recommended, because the return value has multiple meanings.

static class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}

// one important prerequisite is that p and q must be in the whole tree
// the return value has multiple meanings (NOT a good design):
// if null, neither p nor q exists in the tree of root;
// if not null, at least one of p and q exists in the tree of root,
// if p and q exist in the left and right subtree of root respectively, return root itself.
// because in this case, root is the lowest common ancestor
private TreeNode dfs(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) {
        return root;
    }

    TreeNode leftResult = dfs(root.left, p, q);
    TreeNode rightResult = dfs(root.right, p, q);

    if (leftResult != null && rightResult != null) { // root is the lowest common ancestor
        return root;
    } else if (leftResult != null) {
        return leftResult;
    } else if (rightResult != null) {
        return rightResult;
    } else {
        return null;
    }
}

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    return dfs(root, p, q);
}