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236. Lowest Common Ancestor of a Binary Tree

Description

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.

Solutions

The LCA of two nodes p and q in a binary tree is the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).

The idea behind the solution is to recursively search for the nodes p and q in the left and right subtrees. There are a few cases to consider:

  • If either p or q matches the current node, this node must be part of the LCA.
  • If p and q are found in different subtrees of a node, this node is their LCA.

  • If p and q are found in the same subtree, continue searching in that subtree for the LCA.

  • Base Case: If the current node is None, or if it matches either p or q, return the current node. This means we have found one of the nodes we’re looking for, or we’ve reached the end of a path without finding either, in which case we return None.

  • Recursive Search: The function recursively searches the left and right subtrees for p and q.

  • Identifying LCA:
    • If both left and right search calls return non-null values, it means we’ve found p and q in different subtrees of the current node. Hence, the current node is the LCA.
    • If only one of the search calls returns a non-null value, it means both p and q are located in the same subtree, or only one of the nodes was found. Return the non-null result.
  • /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) return root;
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);
        if (left == null) return right;
        if (right == null) return left;
        return root;
    }
}

  • /**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root || root == p || root == q) return root;
        TreeNode* left = lowestCommonAncestor(root->left, p, q);
        TreeNode* right = lowestCommonAncestor(root->right, p, q);
        if (left && right) return root;
        return left ? left : right;
    }
};

  • # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(
        self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode'
    ) -> 'TreeNode':
        if root is None or root == p or root == q:
            return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        return root if left and right else (left or right)

############

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
  def lowestCommonAncestor(self, root, p, q):
    """
    :type root: TreeNode
    :type p: TreeNode
    :type q: TreeNode
    :rtype: TreeNode
    """

    if not root:
      return root

    left = self.lowestCommonAncestor(root.left, p, q)
    right = self.lowestCommonAncestor(root.right, p, q)

    if left and right:
      return root

    if root == p or root == q:
      return root

    if left:
      return left
    if right:
      return right
    return None


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	if root == nil || root == p || root == q {
		return root
	}
	left := lowestCommonAncestor(root.Left, p, q)
	right := lowestCommonAncestor(root.Right, p, q)
	if left == nil {
		return right
	}
	if right == nil {
		return left
	}
	return root
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function lowestCommonAncestor(
    root: TreeNode | null,
    p: TreeNode | null,
    q: TreeNode | null,
): TreeNode | null {
    const find = (root: TreeNode | null) => {
        if (root == null || root == p || root == q) {
            return root;
        }
        const left = find(root.left);
        const right = find(root.right);
        if (left != null && right != null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        return right;
    };
    return find(root);
}


  • /**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function (root, p, q) {
    if (!root || root == p || root == q) return root;
    const left = lowestCommonAncestor(root.left, p, q);
    const right = lowestCommonAncestor(root.right, p, q);
    if (!left) return right;
    if (!right) return left;
    return root;
};


  • // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn find(
        root: &Option<Rc<RefCell<TreeNode>>>,
        p: &Option<Rc<RefCell<TreeNode>>>,
        q: &Option<Rc<RefCell<TreeNode>>>
    ) -> Option<Rc<RefCell<TreeNode>>> {
        if root.is_none() || root == p || root == q {
            return root.clone();
        }
        let node = root.as_ref().unwrap().borrow();
        let left = Self::find(&node.left, p, q);
        let right = Self::find(&node.right, p, q);
        match (left.is_some(), right.is_some()) {
            (true, false) => left,
            (false, true) => right,
            (false, false) => None,
            (true, true) => root.clone(),
        }
    }

    pub fn lowest_common_ancestor(
        root: Option<Rc<RefCell<TreeNode>>>,
        p: Option<Rc<RefCell<TreeNode>>>,
        q: Option<Rc<RefCell<TreeNode>>>
    ) -> Option<Rc<RefCell<TreeNode>>> {
        Self::find(&root, &p, &q)
    }
}

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