235. Lowest Common Ancestor of a Binary Search Tree && 236. Lowest Common Ancestor of a Binary Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

 

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 (M) Lowest Common Ancestor of a Binary Tree

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        int min = Math.min(p.val,q.val);
        int max = Math.max(p.val,q.val);
        
        while(root!=null)
        {
            if(min <= root.val && max >= root.val)
                return root;
            if(min > root.val && max > root.val)
                root = root.right;
            if(min < root.val && max < root.val)
                root = root.left;
        }
        
        return root;   
    }
}

 

 

236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes(REFERENCE EQUAL) in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

 

Solution 1: Recursion

LCA function returns:

1. The LCA if both nodes exist in the tree
2. If last step is not true, then the only node that exists in the tree
3. If last step is not true, then Null (both nodes don’t exist in the tree)

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public 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 && right != null)   return root;
        return left != null ? left : right;
    }
}

 

 

Solution 2: Get the path for each node and compare to get the LCA.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        Deque<TreeNode> pPath = getPath(root, p);
        Deque<TreeNode> qPath = getPath(root, q);
        
        TreeNode lca = null;
        while(!pPath.isEmpty() && !qPath.isEmpty())
        {
            TreeNode pLast = pPath.removeLast();
            TreeNode qLast = qPath.removeLast();
            if(pLast!=qLast)
                break;
            lca = pLast;
        }
                
        return lca;
    }
    
    private Deque<TreeNode> getPath(TreeNode root, TreeNode t)
    {
        Deque<TreeNode> path = new ArrayDeque<TreeNode>();
        Deque<TreeNode> unprocessed = new ArrayDeque<TreeNode>();
        unprocessed.push(root);
        
        while(!unprocessed.isEmpty())
        {
            TreeNode current = unprocessed.pop();
            path.push(current);    
            if(current == t)
            {
                return path;
            }
            
            if(current.right != null)
                unprocessed.push(current.right);
                    
            if(current.left != null)
                unprocessed.push(current.left);
            else
                while(path.peek().right != unprocessed.peek())
                {
                    path.pop();
                }
        }
        return null;
    }
}