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

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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);            }        }    }}

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