Lowest Common Ancestor of a Binary Tree
Given a binary tree and two nodes p and q, return their lowest common ancestor (LCA). The LCA is the deepest node that is an ancestor of both p and q. A node is considered an ancestor of itself.
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Problem
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
Input
The root of a binary tree and two tree nodes `p` and `q`.
Output
The LCA node.
Examples
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Node 3 is the LCA of nodes 5 and 1.
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Node 5 is an ancestor of node 4, so 5 is its own LCA with 4.
The brute-force approach
For each node, find the path from root to p and the path from root to q. Find the deepest node common to both paths.
def find_path(root, target, path):
if not root: return False
path.append(root)
if root == target: return True
if find_path(root.left, target, path) or find_path(root.right, target, path):
return True
path.pop()
return False
path_p, path_q = [], []
find_path(root, p, path_p)
find_path(root, q, path_q)
# Find last common element
lca = None
for a, b in zip(path_p, path_q):
if a == b: lca = a
return lcaO(n) but with two separate DFS passes and O(n) space for paths. The recursive single-pass approach below is cleaner.
Spotting the pattern
This is a Tree Traversal problem. The key question to ask yourself:
If left subtree DFS returns node A and right subtree DFS returns node B (both non-null), what does that tell you about the LCA?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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