Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/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.

解题思路：

CC150里4.7，但是原题是任意二叉树，这里更为简单，被限定为二叉搜索树。

根据BST的性质，root.left.val < root.val < root.right.val，并且所有子树也都是BST。我们的思路就变成判断p和q到底是在root的哪边。

1. 如果都小于root.val，一定同时在左侧，root不是lca，递归搜索左子树。

2. 如果都大于root.val，递归搜索右子树。

3. 否则，分别在root两侧，root就是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(root == null) {
            return null;
        }
        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;
        }
    }
}

 

上面算法的时间复杂度是O(logn)，等于树的高度。下面的算法可以用在任意二叉树，不一定是BST。

仅在root左右子树都分别找到p和q的情况下返回root，说明root是p和q的lca。否则找不到的返回null，找到的一侧返回p或者q。

/**
 * 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(root == null) {
            return null;
        }
        if(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 left == null ? right : left;
        }
        return root;
    }
}

上面的解法时间复杂度为O(n)。