【Leetcode】Lowest Common Ancestor of a Binary Search Tree

【題目】

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.

【思路】

開始一看題。。我去。。懵了，感覺這個easy的題目猛然並麼有什麼思路。看了discuss裏面大神們的解答又頓時恍然大悟了。

首先注意是BST！！說明什麼。。sorted. 左子樹全是比根小得，右子樹全是比根大的。

先判斷，p,q是不是同一顆樹上的，如果不是同一個樹上的，說明一個比根小，一個比根大，那他們不會有比root更low的相同點了，就是root了！

如果是同一個樹上的，比如兩者都比root小，那麼他們的更low的root就在左子樹上。如果都比根大，那麼就

【代碼】

return ((p.val-root.val)*(q.val-root.val)<=0) ? root : lowestCommonAncestor(p.val>root.val?root.right:root.left, p, q);

正常版本。。。

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

             return root;
         }
    }
}