leetcode - 1650. Lowest Common Ancestor of a Binary Tree III_leecode 1650

Description

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;
}
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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.
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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.
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Example 3:

Input: root = [1,2], p = 1, q = 2
Output: 1
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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.
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Solution

Optimized

Find the height difference between p and q, and let the lower node go up height_diff steps first, then both nodes go up. The first common node is their lowest common ancestor.

Time complexity: $o(\log n)$
Space complexity: $o(1)$

Another way of coding this, let’s say p is the lower node, when q reaches the root, we can move q to original p, and when p reaches the root, q is height_diff above the original p, so we can move p to original q again, and now these 2 nodes move together.

Brute force

Find the ancestors of p first, then find the ancestor of q, if the node appears in p’s ancestors, then return.

Time complexity: $o(\log n)$
Space complexity: $o(n)$

Code

Optimized

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        n1, n2 = p, q
        diff = 0
        while n1 or n2:
            if n1 and n2:
                n1 = n1.parent
                n2 = n2.parent
            elif n1:
                n1 = n1.parent
                diff -= 1
            elif n2:
                n2 = n2.parent
                diff += 1
        while diff < 0:
            p = p.parent
            diff += 1
        while diff > 0:
            q = q.parent
            diff -= 1
        while p != q:
            p = p.parent
            q = q.parent
        return p
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Another way of coding:

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        n1, n2 = p, q
        while n1 != n2:
            n1 = n1.parent if n1 else q
            n2 = n2.parent if n2 else p
        return n1
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Brute Force

"""
# Definition for a Node.
class Node:
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.parent = None
"""

class Solution:
    def lowestCommonAncestor(self, p: 'Node', q: 'Node') -> 'Node':
        p_ans = []
        while p:
            p_ans.append(p)
            p = p.parent
        while q:
            if q in p_ans:
                return q
            q = q.parent
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