Algorithm

LeetCode/Lowest Common Ancester Of A Binary Search Tree Or A Binary Tree

Problem Summary

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

  2. What if the tree is just a binary tree ?

Solution

Subproblem 1 is trivial if we know the definition of BST.

Subproblem 2 needs more thinking. There are two ways to solve it.

  1. The first is traditional. We define a variant LCA and a function dfs(TreeNode *ROOT,TreeNode *p,TreeNode *q). dfs returns the number of given nodes that are in the subtree with ROOT as root. If dfs returns 2 and LCA have not been changed, then the answer is ROOT.

  2. Let us define a recursive function f(TreeNode *ROOT,TreeNode *p,TreeNode *q).

    • If the subtree with ROOT as root contains both p and q, f returns the LCA of p and q.
    • If the subtree only contains one of them, f returns that one.
    • If the subtree contains neither of them, f returns NULL.

For each ROOT, we first check whether it is NULL or equal to p or q. If that is the case, return ROOT. Then we call f(ROOT->left,p,q) and f(ROOT->right,p,q), and store the results in LEFT and RIGHT. If they are both not NULL, we know that both subtrees contain one of p and q. Thus ROOT is the LCA we are looking for. If only one of them are not NULL, then return it. If both are NULL, return NULL.

Code 1 — LCA of a binary search tree

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class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
  	    if (root == NULL || p == NULL || q == NULL)
  	      	return NULL;
  	    if (root->val > p->val && root->val > q->val)
  	    	return lowestCommonAncestor(root->left,p,q);
  	    if (root->val < p->val && root->val < q->val)
  	    	return lowestCommonAncestor(root->right,p,q);
  	    return root;
    }
};

Code 2 — LCA of a binary tree, Solution 1

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 // struct TreeNode {
 //    int val;
 //     TreeNode *left;
 //    TreeNode *right;
 //    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 // };
 
class Solution {
public:
	TreeNode *LCA;
	int dfs(TreeNode *root,TreeNode *p,TreeNode *q)
	{
		if (root == NULL)
			return 0;
		int count = 0;
		if (p == root)
			++ count;
		if (q == root)
			++ count;
		count += dfs(root->left,p,q);
		if (count == 2)
		{
			if (!LCA)
				LCA = root;
			return 2;
		}
		
		count += dfs(root->right,p,q);
		if (count == 2 && !LCA)
			LCA = root;
		return count;
	}
    TreeNode* lowestCommonAncestor(TreeNode* root,TreeNode *p,TreeNode *q) {
    	LCA = NULL;
        dfs(root,p,q);
        return LCA;
    }
};

Code 3 – LCA of a binary tree, Solution 2

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 // 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 == NULL || root == p || root == q)
        	return root;
        TreeNode *l = lowestCommonAncestor(root->left,p,q);
        TreeNode *r = lowestCommonAncestor(root->right,p,q);
        if (l && r)
            return root;
	return l ? l:r;
       // return l ? root->left:root->right;
    }
};