[LeetCode 235] Lowest Common Ancestor of a Binary Search 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.

Solution:

1use property of BST

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if((root.val>=p.val && root.val<=q.val)||(root.val<=p.val && root.val>=q.val))
            return root;
        if(root.val>=p.val && root.val>=q.val){
            return lowestCommonAncestor(root.left, p, q);
        }
        return lowestCommonAncestor(root.right, p, q);
    }

Add check to verify if p and q is valid tree node

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null) return null;
        int max = root.val;
        int min = root.val;
        TreeNode t1 = root;
        while(t1!=null){
            min = t1.val;
            t1 = t1.left;
        }
        t1 = root;
        while(t1!=null){
            max = t1.val;
            t1 = t1.right;
        }
        if(p.val>q.val && (min>q.val || max<p.val)) return null;
        if(q.val>p.val && (min>p.val || max<q.val)) return null;
        if((root.val>=p.val && root.val<=q.val)||(root.val<=p.val && root.val>=q.val))
            return root;
        if(root.val>=p.val && root.val>=q.val){
            return lowestCommonAncestor(root.left, p, q);
        }
        return lowestCommonAncestor(root.right, p, q);
    }