Question

236. Lowest Common Ancestor of a Binary Tree

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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

Solution 1: Recursion

We recursively traverse the binary tree:

If the current node is null or equals to $p$ or $q$, then we return the current node;

Otherwise, we recursively traverse the left and right subtrees, and record the returned results as $left$ and $right$. If both $left$ and $right$ are not null, it means that $p$ and $q$ are in the left and right subtrees respectively, so the current node is the nearest common ancestor; If only one of $left$ and $right$ is not null, we return the one that is not null.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

# 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 in (None, p, 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.
 * 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;
    }
    var left = lowestCommonAncestor(root.left, p, q);
    var right = lowestCommonAncestor(root.right, p, q);
    if (left != null && right != null) {
      return root;
    }
    return left == null ? right : left;
  }
}

/**
 * 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 == nullptr || root == p || root == q) {
      return root;
    }
    auto left = lowestCommonAncestor(root->left, p, q);
    auto right = lowestCommonAncestor(root->right, p, q);
    if (left && right) {
      return root;
    }
    return left ? left : right;
  }
};

/**
 * 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 && right != nil {
    return root
  }
  if left != nil {
    return left
  }
  return right
}

/**
 * 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 {
  if (!root || root === p || root === q) {
    return root;
  }
  const left = lowestCommonAncestor(root.left, p, q);
  const right = lowestCommonAncestor(root.right, p, q);
  return left && right ? root : left || right;
}

// 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 {
  pub fn lowest_common_ancestor(
    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;
    }
    let left = Self::lowest_common_ancestor(
      root.as_ref().unwrap().borrow().left.clone(),
      p.clone(),
      q.clone()
    );
    let right = Self::lowest_common_ancestor(
      root.as_ref().unwrap().borrow().right.clone(),
      p.clone(),
      q.clone()
    );
    if left.is_some() && right.is_some() {
      return root;
    }
    if left.is_none() {
      return right;
    }
    return left;
  }
}

/**
 * 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);
  return left && right ? root : left || right;
};