【LeetCode】235 Lowest Common Ancestor of a Binary Search Tree

Lowest Common Ancestor of a Binary Search Tree

Total Accepted: 3808 Total Submissions: 9820 My Submissions Question Solution

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.

【解题思路】

二叉搜索树，根据特点搜索，比当前值大，搜索右子树，比当前值小，搜索左子树，相等就返回结果。

Java AC

/**
 * 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) {
        List<TreeNode> list1 = new ArrayList<TreeNode>();
        List<TreeNode> list2 = new ArrayList<TreeNode>();
        getPath(root, p, list1);
        getPath(root, q, list2);
        int size1 = list1.size();
        int size2 = list2.size();
        int len = size1 < size2 ? size1 : size2;
        int k = 0;
        while (k < len){
            TreeNode node1 = list1.get(k);
            TreeNode node2 = list2.get(k);
            if (node1 != node2){
                break;
            }
            k++;
        }
        return list1.get(k - 1);
    }
    
    private void getPath(TreeNode root, TreeNode d, List<TreeNode> list){
        if (root != null){
            int rval = root.val;
            int dval = d.val;
            list.add(root);
            if (dval < rval){
                getPath(root.left, d, list);
            }else if (dval > rval){
                getPath(root.right, d, list);
            }
        }
    }
}