Lowest Common Ancestor of a Binary Tree IV

Asked atJ.P. MorganMicrosoft
1Give yourself 5 minutes
2Answer out loud, not in your head
3Then compare with the answer below

The problem

Given the root of a binary tree and an array of TreeNodes objects nodes, return the** lowest common ancestor (LCA) **of all the nodes in nodes.

All the nodes will exist in the tree, and all values of the tree's nodes are unique.

The lowest common ancestor of n nodes p1,p2,...pn in a binary tree T is the lowest node that has every pi as a descendant (where we allow a node to be a descendant of itself) for every valid i. A descendant of a node x is a node y that is on the path from node x to some leaf node.

Input: root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [4,7] Output: 2 **Explanation: ** The lowest common ancestor of nodes 4 and 7 is node 2.

**Input: **root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [1] Output: 1 Explanation: The lowest common ancestor of a single node is the node itself.

**Input: **root = [3,5,1,6,2,0,8,null,null,7,4], nodes = [7,6,2,4]

  • The number of nodes in the tree is in the range [1, 104].
  • -109 <= Node.val <= 109
  • All Node.val are unique.
  • All nodes[i] will exist in the tree.
  • All nodes[i] are distinct.

cpp

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     int data;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : data(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, vector<TreeNode*> &nodes) {
        //Your Code Goes Here
    }
};

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int data;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { data = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, List<TreeNode> nodes) {
        // Your Code Goes Here
    }
}

python

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

class Solution:
    def lowestCommonAncestor(self, root, nodes):

javascript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *      constructor(data = 0, left = null, right = null){
 *          this.data = data;
 *          this.left = null;
 *          this.right = null;
 *      }
 * }
 **/

class Solution {
    lowestCommonAncestor(root,nodes) {
        //your code goes here
    }
}

csharp

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int data;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val) { data = val; }
 * }
 */
public class Solution {
    public TreeNode LowestCommonAncestor(TreeNode root, List<TreeNode> nodes) {
        // Your Code Goes Here
    }
}

go

func lowestCommonAncestor(root *TreeNode, nodes []*TreeNode) *TreeNode {
    // Your code goes here
}
Stuck? Show a way to structure it+
  1. 01Put all target-node references in a set.
  2. 02Run postorder DFS.
  3. 03Return a target immediately when encountered.
  4. 04If two child calls are non-null, return the current node.
  5. 05Otherwise propagate the non-null child result.

Reference answer

Then expect these follow-ups

  • How would you validate that every requested node occurs?

    Tests: robustness

  • Can this be iterative?

    Tests: parent maps

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