LeetCode-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.

思路，先判断入口是否有非法输入。

1 如果某一个root==p || root == q，那么LCA肯定是root（因为是top down，LCA肯定在root所囊括的树上，而root又是p q其中一个节点了，那么另外一个节点肯定在root之下，那么root就是LCA），那么返回root

2 如果root<min(p, q)，那么LCA肯定在右子树上，那么递归

3 如果max(p, q)<root，那么LCA肯定在左子树上，那么递归

4 如果p<root<q，那么root肯定为LCA

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        /*
        if(p == root || q == root || root == null){
            return root;
        }
        TreeNode left=lowestCommonAncestor(root.left, p, q);
        TreeNode right=lowestCommonAncestor(root.right, p, q);
        //在两边
        if(left != null && right != null ) return root;
        //在一边
        if(left != null){
            
        }
        */
        if(root==null || root == p || root == q){
            return root;
        }
        if(root.val < Math.min(p.val, q.val)){
            return lowestCommonAncestor(root.right, p, q);
        }
        else if(root.val > Math.max(p.val, q.val)){
            return lowestCommonAncestor(root.left, p, q);
        }
        else {
            return root;
        }
    }
}

　　二刷：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root == null){
            return null;
        }
        if((root.val >= p.val && root.val <= q.val) || (root.val <= p.val && root.val >= q.val)){
            return root;
        }
        else if(root.val < p.val && root.val < q.val){
            return lowestCommonAncestor(root.right, p, q);
        }
         else{
            return lowestCommonAncestor(root.left, p, q);
        }
       
    }
}