题目描述

题目链接：236. Lowest Common Ancestor of a Binary Tree

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

例子

例子 1

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

例子 2

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

例子 3

Input: root = [1,2], p = 1, q = 2
Output: 1

Constraints

• The number of nodes in the tree is in the range [2, 10^5].
• -10^9 <= Node.val <= 10^9
• All Node.val are unique.
• p != q
• p and q will exist in the BST.

解题思路

这里涉及到在二叉树中对节点的搜索，思路如下：

• 对一棵树同时搜索 p 或者 q
• 如果根结点是空，直接返回空指针
• 如果根结点是 p 或者 q，直接返回根结点（因为假如另一个节点在同一棵树内的话（题目保证了两个节点都会存在），最深的祖先节点一定是根结点）
• 否则对左右两颗子树进行同样操作，得到返回结果 left 和 right
  • 假如 left 和 right 都非空，表示在左子树和右子树都搜到了其中一个节点，此时根结点为最近祖先节点，返回根结点
  • 假如 left 和 right 只有一个非空节点，表示除了该非空节点的子树外，其他节点都不包含另一个节点，表明另一个节点一定在该节点的子树下，返回该节点作为最近祖先节点
  • 假如 left 和 right 都为节点：不可能，因为题目保证两个节点都存在

代码如下：

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/**
 * 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 (root == nullptr) return nullptr;
        if (root == p || root == q) return root;
        // search node in left subtree
        TreeNode* left_search = lowestCommonAncestor(root->left, p, q);
        TreeNode* right_search = lowestCommonAncestor(root->right, p, q);
        // we find node in both subtree
        if (left_search && right_search) {
            return root;
        } else if (left_search) {
            return left_search;
        } else {
            return right_search;
        }
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(h) — 递归深度

GitHub 代码同步地址： 236.LowestCommonAncestorOfABinaryTree.cpp

其他题目：
GitHub: Leetcode-C++-Solution
博客： Leetcode-Solutions