235. Lowest Common Ancestor of a Binary Search Tree

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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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Note:

• All of the nodes' values will be unique.

• p and q are different and both values will exist in the BST.

Solution

從root開始透過找 看是否有節點.val符合 小於等於 maxOf(p.val, q.val), 大於minOf(p.val, q.val)即為解，

應該只會重複tree的高度， worst case would be O(N)， average case would be O(logN)

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