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1644. Lowest Common Ancestor of a Binary Tree II πŸ”’

Description

Given the root of a binary tree, return the lowest common ancestor (LCA) of two given nodes, p and q. If either node p or q does not exist in the tree, return null. All values of the nodes in the tree are unique.

According to the definition of LCA on Wikipedia: "The lowest common ancestor of two nodes p and q in a binary 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)". A descendant of a node x is a node y that is on the path from node x to some leaf node.

 

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. A node can be a descendant of itself according to the definition of LCA.

Example 3:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 10
Output: null
Explanation: Node 10 does not exist in the tree, so return null.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • p != q

 

Follow up: Can you find the LCA traversing the tree, without checking nodes existence?

Solutions

Solution 1

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# 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':
        def dfs(root, p, q):
            if root is None:
                return False
            l = dfs(root.left, p, q)
            r = dfs(root.right, p, q)
            nonlocal ans
            if l and r:
                ans = root
            if (l or r) and (root.val == p.val or root.val == q.val):
                ans = root
            return l or r or root.val == p.val or root.val == q.val

        ans = None
        dfs(root, p, q)
        return ans

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode ans;

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        dfs(root, p, q);
        return ans;
    }

    private boolean dfs(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return false;
        }
        boolean l = dfs(root.left, p, q);
        boolean r = dfs(root.right, p, q);
        if (l && r) {
            ans = root;
        }
        if ((l || r) && (root.val == p.val || root.val == q.val)) {
            ans = root;
        }
        return l || r || root.val == p.val || root.val == q.val;
    }
}

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/**
 * 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) {
        dfs(root, p, q);
        return ans;
    }

private:
    TreeNode* ans = nullptr;

    bool dfs(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (!root) {
            return false;
        }
        bool l = dfs(root->left, p, q);
        bool r = dfs(root->right, p, q);
        if (l && r) {
            ans = root;
        }
        if ((l || r) && (root->val == p->val || root->val == q->val)) {
            ans = root;
        }
        return l || r || root->val == p->val || root->val == q->val;
    }
};

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/**
 * 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) {
    const dfs = root => {
        if (!root) {
            return false;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        if (l && r) {
            ans = root;
        }
        if ((l || r) && (root.val === p.val || root.val === q.val)) {
            ans = root;
        }
        return l || r || root.val === p.val || root.val === q.val;
    };
    let ans = null;
    dfs(root);
    return ans;
};

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