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.

//二叉搜索树有个性质：左子树的值都比根节点小，右子树的值比根节点大。那么我们的思路就是递归比较；
//如果输入的两个节点的值比当前节点小，说明是在当前根节点的左子树中；反之则在右子树中。如果当前根节点比一个大，比一个小，那么第一个出现的这样的节点就是最近的父亲节点了。

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

//另外还可以无视BST的性质，当成是BT来做，自下而上遍历最后看是否遍历到p，q节点

public class Solution {
    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;
        }
        if(left !=null){
            return left;
        }
        if(right!=null){
            return right;
        }
        return null;
    }
}