1650. Lowest Common Ancestor of a Binary Tree III

Given two nodes of a binary tree p and q, return their lowest common ancestor (LCA).

Each node will have a reference to its parent node. The definition for Node is below:

class Node {
    public int val;
    public Node left;
    public Node right;
    public Node parent;
}

According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p and q in a tree T is the lowest node 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 exist in the tree.

Analysis

Since every node carries a pointer straight up to its parent, walking from p (or q) to the root traces out a path, and this problem reduces exactly to Intersection of Two Linked Lists (LeetCode 160): find the first node where the p -> root path and the q -> root path merge.

The classic trick for that problem works here too, without needing to compute either path's length up front: walk two pointers a starting at p and b starting at q, one step at a time. Whenever a pointer runs off the top (hits a null parent), redirect it to start over from the other node instead of null. Because both pointers together cover a combined distance equal to depth(p) + depth(q) before they first coincide, this rerouting exactly compensates for the difference in depth between p and q, so a and b are guaranteed to meet at the LCA.

• Time: $O(h)$, where $h$ is the height of the tree (bounded by the combined depth of p and q).
• Space: $O(1)$, since we only keep two pointers, unlike the hash-set approach which needs $O(h)$ space to record one path.

Code

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* parent;
};
*/

class Solution {
public:
    Node* lowestCommonAncestor(Node* p, Node* q) {
        Node* a = p;
        Node* b = q;
        while (a != b) {
            a = a->parent ? a->parent : q;
            b = b->parent ? b->parent : p;
        }
        return a;
    }
};