Skip to content

236 Lowest Common Ancestor of a Binary Tree

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).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

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 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.

Note:

All of the nodes’ values will be unique. p and q are different and both values will exist in the binary tree.

Solution 1: Recursion

Time complexity: O(n)

Space complexity: O(h)

For a given root, recursively call LCA(root.left, p, q) and LCA(root.right, p, q)

if both returns a valid node which means p, q are in different subtrees, then root will be their LCA.

if only one valid node returns, which means p, q are in the same subtree, return that valid node as their LCA.

class Solution {
  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if (root == null || root == p || root == q) return root;
    TreeNode l = lowestCommonAncestor(root.left, p, q);
    TreeNode r = lowestCommonAncestor(root.right, p, q);
    if (l == null || r == null) return l == null ? r : l;
    return root;
  }
}

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

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if (root == p || root == q) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        else {
            return right;
        }
    }
}
    1. Same Tree
    1. Smallest Subtree with all the Deepest Nodes