所有的LeetCode题解索引，可以看这篇文章——【算法和数据结构】LeetCode题解。
一、题目
二、解法
思路分析： 根据定义，最近祖先节点需要遍历节点的左右子树，然后才能知道是否为最近祖先节点。那么这种需求是先遍历左右节点，然后遍历中间节点，属于左右中的后序遍历模式。因此在程序当中，我们选择递归中序遍历。输入参数为根节点p q节点。终止条件是当前节点和p q当中任意一个节点相等时就返回，遍历到空节点也返回。因为是后序遍历，根据遍历完左右子树后的返回值确定返回参数，如果返回值都不为空，则当前节点就是最近祖先节点。如果left为空，right不为空，则最近祖先节点在右子树，反之亦然。均为空则返回NULL。
程序如下：
class Solution2 {
public:
// 后序遍历: 左右中
// 1、输入参数
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 2、终止条件
if (root == q || root == p || root == NULL) return root; // 如果节点相等或者是空节点返回
// 3、单层递归逻辑
TreeNode* left = lowestCommonAncestor(root->left, p, q); // 左
TreeNode* right = lowestCommonAncestor(root->right, p, q); // 右
// 1、返回值
if (left != NULL && right != NULL) return root;
if (left == NULL && right != NULL) return right;
else if (left != NULL && right == NULL) return left;
else { // (left == NULL && right == NULL)
return NULL;
}
}
};
代码优化：
class Solution {
public:
// 后序遍历: 左右中
// 1、输入参数
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 2、终止条件
if (root == p || root == q || root == NULL) return root; // 如果节点相等或者是空节点返回
// 3、单层递归逻辑
TreeNode* left = lowestCommonAncestor(root->left, p, q); // 左
TreeNode* right = lowestCommonAncestor(root->right, p, q); // 右
if (left != NULL && right != NULL) return root;
if (left == NULL) return right;
// 1、返回值
return left;
}
};
三、完整代码
# include <iostream>
# include <vector>
# include <string>
# include <queue>
using namespace std;
// 树节点定义
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode* left, TreeNode* right) : val(x), left(left), right(right) {}
};
class Solution {
public:
// 后序遍历: 左右中
// 1、输入参数
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 2、终止条件
if (root == p || root == q || root == NULL) return root; // 如果节点相等或者是空节点返回
// 3、单层递归逻辑
TreeNode* left = lowestCommonAncestor(root->left, p, q); // 左
TreeNode* right = lowestCommonAncestor(root->right, p, q); // 右
if (left != NULL && right != NULL) return root;
if (left == NULL) return right;
// 1、返回值
return left;
}
};
class Solution2 {
public:
// 后序遍历: 左右中
// 1、输入参数
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
// 2、终止条件
if (root == q || root == p || root == NULL) return root; // 如果节点相等或者是空节点返回
// 3、单层递归逻辑
TreeNode* left = lowestCommonAncestor(root->left, p, q); // 左
TreeNode* right = lowestCommonAncestor(root->right, p, q); // 右
// 1、返回值
if (left != NULL && right != NULL) return root;
if (left == NULL && right != NULL) return right;
else if (left != NULL && right == NULL) return left;
else { // (left == NULL && right == NULL)
return NULL;
}
}
};
// 前序遍历迭代法创建二叉树，每次迭代将容器首元素弹出（弹出代码还可以再优化）
void Tree_Generator(vector<string>& t, TreeNode*& node) {
if (!t.size() || t[0] == "NULL") return; // 退出条件
else {
node = new TreeNode(stoi(t[0].c_str())); // 中
if (t.size()) {
t.assign(t.begin() + 1, t.end());
Tree_Generator(t, node->left); // 左
}
if (t.size()) {
t.assign(t.begin() + 1, t.end());
Tree_Generator(t, node->right); // 右
}
}
}
template<typename T>
void my_print(T& v, const string msg)
{
cout << msg << endl;
for (class T::iterator it = v.begin(); it != v.end(); it++) {
cout << *it << ' ';
}
cout << endl;
}
template<class T1, class T2>
void my_print2(T1& v, const string str) {
cout << str << endl;
for (class T1::iterator vit = v.begin(); vit < v.end(); ++vit) {
for (class T2::iterator it = (*vit).begin(); it < (*vit).end(); ++it) {
cout << *it << ' ';
}
cout << endl;
}
}
// 层序遍历
vector<vector<int>> levelOrder(TreeNode* root) {
queue<TreeNode*> que;
if (root != NULL) que.push(root);
vector<vector<int>> result;
while (!que.empty()) {
int size = que.size(); // size必须固定, que.size()是不断变化的
vector<int> vec;
for (int i = 0; i < size; ++i) {
TreeNode* node = que.front();
que.pop();
vec.push_back(node->val);
if (node->left) que.push(node->left);
if (node->right) que.push(node->right);
}
result.push_back(vec);
}
return result;
}
// 前序遍历,找二叉树中指定的键值
TreeNode* traversal_preOrder(TreeNode* cur, int val) {
if (cur == NULL) return NULL;
if (cur->val == val) return cur; // 中
if (traversal_preOrder(cur->left, val) != NULL) return traversal_preOrder(cur->left, val); // 左
if (traversal_preOrder(cur->right, val) != NULL) return traversal_preOrder(cur->right, val); // 右
return NULL;
}
int main()
{
// 构建二叉树
vector<string> t = { "3", "5", "6", "NULL", "NULL", "2", "7", "NULL", "NULL", "4", "NULL", "NULL", "1", "0", "NULL", "NULL", "8", "NULL", "NULL"}; // 前序遍历
my_print(t, "目标树");
TreeNode* root = new TreeNode();
Tree_Generator(t, root);
vector<vector<int>> tree = levelOrder(root);
my_print2<vector<vector<int>>, vector<int>>(tree, "目标树:");
// 构建p, q节点
int p = 5, q = 1;
TreeNode* P_node = traversal_preOrder(root, p);
TreeNode* Q_node = traversal_preOrder(root, q);
// 找最近公共祖先节点
Solution s;
TreeNode* result = s.lowestCommonAncestor(root, P_node, Q_node);
cout << "节点 " << P_node->val << " 和节点 " << Q_node->val << " 的最近公共祖先节点为 " << result->val << endl;
system("pause");
return 0;
}
end
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