题解 | #在二叉树中找到两个节点的最近公共祖先#

import java.util.*;
//采用递归方法，二叉树解题的常用方法
/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param root TreeNode类 
     * @param o1 int整型 
     * @param o2 int整型 
     * @return int整型
     */
    public int lowestCommonAncestor (TreeNode root, int o1, int o2) {
        // write code here
        return leastCommonAncestors(root,o1,o2).val;
    }
    
    public TreeNode leastCommonAncestors(TreeNode root, int o1, int o2){
        //递归出口，空指针
        if(root == null){
            return null;
        }
        //递归出口
        if(root.val == o1 || root.val == o2){
            return root;
        }
        //递归左子树
        TreeNode left = leastCommonAncestors(root.left,o1,o2);
        //递归右子树
        TreeNode right = leastCommonAncestors(root.right,o1,o2);
        //如果left为空,则说明两个节点在root的右子树上
        if(left == null){
            return right;
        }
        //如果right为空,则说明两个节点在root的左子树上
        if(right == null){
            return left;
        }
        
        //如果left,right都不为空,则说明两个节点在root上
        return root;
    }
}