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本帖最后由 糖逗 于 2020-5-8 18:06 编辑
题目描述:
- 给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。
- 百度百科中最近公共祖先的定义为:“对于有根树 T 的两个结点 p、q,最近公共祖先表示为一个结点 x,满足 x 是 p、q 的祖先且 x 的深度尽可能大(一个节点也可以是它自己的祖先)。”
- 例如,给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]
-
- 示例 1:
- 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
- 输出: 6
- 解释: 节点 2 和节点 8 的最近公共祖先是 6。
- 示例 2:
- 输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
- 输出: 2
- 解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。
-
- 说明:
- 所有节点的值都是唯一的。
- p、q 为不同节点且均存在于给定的二叉搜索树中。
复制代码
- #include <iostream>
- #include <vector>
- #include <string>
- #include<malloc.h>
- #include <math.h>
- using namespace std;
- struct TreeNode{
- int val;
- TreeNode* left;
- TreeNode* right;
- TreeNode(int x): val(x), left(NULL), right(NULL){
- }
- };
- TreeNode* CreateTree(vector<int>& input){
- TreeNode* tree = (TreeNode*)malloc(sizeof(TreeNode)*input.size());
- for(int i = 0; i < input.size() ;i++){
- tree[i].val = input[i];
- tree[i].left = NULL;
- tree[i].right = NULL;
-
- }
- for(int i = 0; i <= input.size()/2-1; i++){
- if(2*i+1 <= input.size()){
- tree[i].left = &tree[2*i+1];
- }
- if(2*i+2<=input.size()){
- tree[i].right = &tree[2*i+2];
- }
- }
- return tree;
- }
- void middle(TreeNode* root, vector<vector<int> >& res, int left, int right, int depth){
- if(root == NULL) return;
- int insert = left + (right - left) / 2;
- res[depth][insert] = root->val;
-
- middle(root->left, res, left, insert - 1, depth + 1);
- middle(root->right, res, insert + 1, right, depth + 1);
- }
- int treeDepth(TreeNode* root){
- if(root == NULL) return 0;
- return max(treeDepth(root->left) + 1, treeDepth(root->right) + 1);
- }
-
- void printTree(TreeNode* root) {
- int depth = treeDepth(root);
- int width = pow(2, depth) - 1;
- vector<vector<int> > res(depth, vector<int>(width, 0));
- middle(root, res, 0, width - 1, 0);
- for(int i = 0; i < res.size(); i++){
- for(int j = 0; j < res[i].size();j++){
- if(res[i][j] == 0){
- cout << " ";
- }
- else{
- cout << res[i][j];
- }
-
- }
- cout << endl;
- }
- cout << "------------------" << endl;
- }
- TreeNode* findNode(TreeNode*root, int number1){
- if(root -> val == number1) return root;
- if(root -> val > number1 && root->left != NULL) return findNode(root -> left, number1);
- if(root -> val < number1 && root->right != NULL) return findNode(root -> right, number1);
- return NULL;
- }
- TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
- if((root->val-p->val)*(root->val-q->val) <= 0) return root;
- if(root -> val < p -> val) return lowestCommonAncestor(root -> right, p, q);
- if(root -> val > p -> val) return lowestCommonAncestor(root -> left, p, q);
- return NULL;
-
- }
- int main(void){
- vector<int>input;
- int number;
- cout << "please send numbers for this tree:" << endl;
- while(cin >> number){
- input.push_back(number);
- }
- cin.clear();
- TreeNode* root = CreateTree(input);
-
- printTree(root);
- cout << "please send the first number that you want to find in this tree" << endl;
- int number1;
- cin >> number1;
- cin.clear();
- TreeNode* find_number1 = findNode(root, number1);
- printTree(find_number1);
-
- cout << "please send the second number that you want to find in this tree" << endl;
- int number2;
- cin >> number2;
- TreeNode* find_number2 = findNode(root, number2);
- printTree(find_number2);
-
- TreeNode* res = lowestCommonAncestor(root, find_number1, find_number2);
- printTree(res);
- return 0;
- }
复制代码
注意事项:
1.这里的代码包括输入vector转化为二叉树、将二叉树打印、寻找二叉树中等于目标值的根节点树、返回两个树之间公共根节点的完整树。
2.这里的输入必须保证根节点介于左右两个节点之间:左节点值<根节点值<右节点值 例如[4,2,8,1,3,7,9]可以,[1,2,3,4,5,6,7]不可以。 |
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