[LeetCode]235. Lowest Common Ancestor of a Binary Search Tree&236. Lowest Common Ancestor of a Binar

235 . Lowest Common Ancestor of a Binary Search Tree
Easy

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.

8ms:

  public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {        int r_v = root.val;        int p_v = p.val;        int q_v = q.val;        if((r_v-p_v)*(r_v-q_v)<=0)            return root;        else            return lowestCommonAncestor(r_v>p_v?root.left:root.right, p, q);    }

236 . Lowest Common Ancestor of a Binary Tree
Medium

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

11ms:

 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;    }