235. Lowest Common Ancestor of a Binary Search Tree(Tree-Easy)

Jack_Cui

发布于 2018-01-08 15:08:06

5650

发布于 2018-01-08 15:08:06

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).”

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.

题目：求二叉树的LCA，也就是两个节点的最低公共祖先。比如上图的二叉树，节点2和8的LCA是6，节点2和4的LCA是2。

思路：观察给定的二叉树可知，二叉树中任何节点：左节点的值 < 根节点的值 < 右节点的值。根据这个性质，可以做出如下判断：

如果p、q都比根节点小，则在左子树中递归查找最低公共祖先节点。
如果p、q都比根节点大，则在右子树中递归查找最低公共祖先节点。
如果p、q一个比根节点大，一个比根节点小，或者有一个等于根节点，则根节点即为最低公共祖先节点。

Language : cpp

递归方法：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if (p->val < root->val && root->val > q->val)
            return lowestCommonAncestor(root->left, p, q);
        else if (p->val > root->val && root->val < q->val)
            return lowestCommonAncestor(root->right, p, q);
        else
            return root;
    }
};

Language : python

递归方法：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if p.val < root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        elif p.val > root.val < q.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

迭代方法：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        while True:
            if p.val < root.val > q.val:
                root = root.left
            elif p.val > root.val < q.val:
                root = root.right
            else:
                return root

代码获取： https://github.com/Jack-Cherish/LeetCode

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原始发表：2017-07-20 ，如有侵权请联系 cloudcommunity@tencent.com 删除

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