leetcode235. Lowest Common Ancestor of a Binary Search Tree

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本文介绍了解决二叉搜索树中寻找两个节点的最低公共祖先(LCA)的问题。提供了三种不同的递归解决方案，包括基本递归方法、优化后的递归方法以及进一步改进的方法。

235. 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.

解法一

如果比root小，查找root左端；如果比root大，查找root右端。

/**
 * 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 root;
        }
        if (root.val < p.val && root.val < q.val) {
            root = root.right;
        } else if (root.val > p.val && root.val > q.val) {
            root = root.left;
        }
        if (root == null) {
            return root;
        }
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (root == p || root == q) {
            return root;
        }
        if (left != null && right !=null) {
            return root;
        }
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        return null; 
    }
}

这里写图片描述

解法二

/**
 * 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) {
        TreeNode result = root;
        if (root.val < p.val && root.val < q.val) {
            result = lowestCommonAncestor(root.right, p, q);
        } else if (root.val > p.val && root.val > q.val) {
            result = lowestCommonAncestor(root.left, p, q);
        } else {
            result = root;
        }
        return result;
    }
}

这里写图片描述

解法三

针对解法一的改进，将相等提前到前面，如果一左一右直接返回root

/**
 * 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 root;
        }
        if (root.val < p.val && root.val < q.val) {
            root = root.right;
        } else if (root.val > p.val && root.val > q.val) {
            root = root.left;
        } else {
            return root;
        }
        if (root == null) {
            return root;
        }
        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 root;
        }
        if (left != null) {
            return left;
        }
        if (right != null) {
            return right;
        }
        return null; 
    }
}