二叉树

1. 树中两个结点的最低公共祖先

原题链接

方法一：公共路径

分别找出根节点到两个节点的路径，则最后一个公共节点就是最低公共祖先了。

时间复杂度分析：需要在树中查找节点，复杂度为O(n)

/*** 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:int findPath(TreeNode*root, TreeNode* p, vector<TreeNode*>&path){if(!root)return 0;if(root->val==p->val){path.push_back(root);return 1;}int l = findPath(root->left,p,path);int r = findPath(root->right,p,path);if(l==1||r==1)path.push_back(root);return l==1||r==1;}TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {vector<TreeNode*>path1,path2;findPath(root,p,path1);findPath(root,q,path2);if(path1.empty()||path2.empty())return NULL;TreeNode* res =NULL;for(int i = 0;i<path1.size();i++){if(i>=path1.size()||i>=path2.size())break;if(path1[path1.size()-1-i]==path2[path2.size()-1-i])res = path1[path1.size()-1-i];elsebreak;}return res;}
};

方法二：递归

考虑在左子树和右子树中查找这两个节点，如果两个节点分别位于左子树和右子树，则最低公共祖先为自己(root)，若左子树中两个节点都找不到，说明最低公共祖先一定在右子树中，反之亦然。考虑到二叉树的递归特性，因此可以通过递归来求得。

时间复杂度分析：需要遍历树，复杂度为 O(n)

class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if(!root)return NULL;if(root==p||root==q)return root;TreeNode* left = lowestCommonAncestor(root->left, p, q);TreeNode* right = lowestCommonAncestor(root->right, p, q);if(left&&right)return root;if(left==NULL)return right;elsereturn left;}
};

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public boolean flag = false;public TreeNode ans = null;public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {dfs(root,p,q);return ans;}public TreeNode dfs(TreeNode root,TreeNode p,TreeNode q){if(root == null){return null;}TreeNode l = dfs(root.left,p,q);TreeNode r = dfs(root.right,p,q);if(l != null && r != null){ans = root;return ans;} else if((l != null || r != null) && (root.val == p.val || root.val == q.val)){ans = root;return ans;} else if(root.val == p.val && root.val == q.val) {ans = root;return ans;} else if(root.val == p.val || root.val == q.val) {return root;} else if((l != null || r != null)){return root;}return null;}}

2. 重建二叉树

原题链接

根据前序遍历和中序遍历 得到树

过程如下：

1. 首先根据前序遍历找到 根节点
2. 找到中序遍历中，该根节点的位置
3. 中序中 位于 根节点左边的就是 左子树，右边的就是右子树
4. 由于我们需要在中序遍历中找到根节点的位置，那么每次都需要遍历中序遍历，不如直接用unordered_map存储数值和位置
5. 便于写代码，我们可以每次把mp[根节点] 的位置 用变量表示出来

本题的代码不需要死记硬背

就需要明白

由前序确定根节点
由中序确定左右子树的个数
由左右子树的个数确定下一个根节点的位置

根据这三点去写代码即可

/*** 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:unordered_map<int,int> pos;TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {int n = preorder.size();for (int i = 0; i < n; i ++ )pos[inorder[i]] = i;return dfs(preorder, inorder, 0, n - 1, 0, n - 1);}TreeNode* dfs(vector<int>&pre, vector<int>&in, int pl, int pr, int il, int ir){if (pl > pr) return NULL;int k = pos[pre[pl]] - il;TreeNode* root = new TreeNode(pre[pl]);root->left = dfs(pre, in, pl + 1, pl + k, il, il + k - 1);root->right = dfs(pre, in, pl + k + 1, pr, il + k + 1, ir);return root;}
};

补充题：树的遍历

#include <iostream>
#include <string>
#include <unordered_map>
#include <vector>
#include <algorithm>
#include<memory>
using namespace std;const  int N = 35;
int n;
int inorder[N], postorder[N];
unordered_map<int, int > leftChile, rightChile;//哈希表保存树，leftChile[i] = j: i 的左儿子是j,rightChilet同理
unordered_map<int, int > h;//保存中序遍历中各节点的位置int dfs(int postorder[], int inorder[], int l1, int r1, int l2, int r2)//构造二叉树
{   if (l1 > r1) return 0;//没有节点，返回0int root = postorder[r1];//根结点为后续遍历的最后一个节点int k = h[root];//找到根节点在序遍历中的位置leftChile[root] = dfs(postorder, inorder, l1, k - 1 - l2 + l1, l2, k - 1);//构造左儿子rightChile[root] = dfs(postorder, inorder,r1-1 - (r2 - (k +1)) , r1 -1, k + 1, r2);//构造右儿子return root;
}int main()
{cin >> n;//输入for (int i = 0; i < n; i++)cin >> postorder[i];for (int i = 0; i < n; i++){cin >> inorder[i];h[inorder[i]] = i;//保存中序遍历中各个节点的位置}int root = dfs(postorder, inorder, 0, n - 1, 0, n - 1);//构造二叉树//数组模拟队列int q[N], hh = 0, tt = -1;//按层次遍历if (root)//非0 表示有节点q[++tt] = root;while (hh <= tt){int t = q[hh++];if (leftChile[t]) q[++tt] = leftChile[t];//非0 为节点，入队列if (rightChile[t]) q[++tt] = rightChile[t];//非0 为节点，入队列}for (int i = 0; i <= tt; i++)//队列中保存的就是按层次遍历的结果cout << q[i] << " ";return 0;
}

3. 二叉树的下一个节点

原题链接

中序遍历：左根右

本题要分析节点的特点

1. 如果节点有右子树，那么右子树的最左边的节点就是该节点后序
2. 如果没有右子树，会有三种可能，在代码中有体现

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode father;*     TreeNode(int x) { val = x; }* }*/
class Solution {/*** 模拟* 时间复杂度：O(height)，height 为二叉树的高度* 空间复杂度：O(1)*/public TreeNode inorderSuccessor(TreeNode p) {TreeNode node = p;// Case 1. 如果该节点有右子树，那么下一个节点就是其右子树中最左边的节点if (node.right != null) {node = node.right;while (node.left != null) {node = node.left;}return node;}if(node.father != null && node.father.left == node)return node.father;if(node.father != null && node.father == null)return null;while(node.father!=null && node.father.right == node){node = node.father;}return node.father;}
}

4. 树的子结构（ 递归中调用递归 ）

原题链接

/*** 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:bool hasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) {if (!pRoot1 || !pRoot2) return false;if (isSame(pRoot1, pRoot2)) return true;return hasSubtree(pRoot1->left, pRoot2) || hasSubtree(pRoot1->right, pRoot2);}bool isSame(TreeNode* pRoot1, TreeNode* pRoot2) {if (!pRoot2) return true;if (!pRoot1 || pRoot1->val != pRoot2->val) return false;return isSame(pRoot1->left, pRoot2->left) && isSame(pRoot1->right, pRoot2->right);}
};

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public boolean hasSubtree(TreeNode pRoot1, TreeNode pRoot2) {// 类比字符串匹配if (pRoot1 == null || pRoot2 == null) return false;if (isPart(pRoot1, pRoot2)) return true;// 如果以 pRoot1 开始的子树不能包含 pRoot2 的话，就从 pRoot1.left 和 pRoot1.right 开始return hasSubtree(pRoot1.left, pRoot2) || hasSubtree(pRoot1.right, pRoot2);}// 以 pRoot1 节点为根的情况下，pRoot1 和 pRoot2 同时对应遍历，如果遍历到的值都相同，且 pRoot2 先遍历到 null 或者同时遍历到 null，说明 pRoot2 是 pRoot1 的子结构public boolean isPart(TreeNode pRoot1, TreeNode pRoot2) {if (pRoot2 == null) return true;if (pRoot1 == null || pRoot1.val != pRoot2.val) return false;// 同时遍历左右子树return isPart(pRoot1.left, pRoot2.left) && isPart(pRoot1.right, pRoot2.right);}
}

5. 二叉树的镜像（两个指针互换可用 swap）

原题链接

/*** 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:void mirror(TreeNode* root) {if (!root) return;swap(root->left, root->right);mirror(root->left);mirror(root->right);}
};

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public void mirror(TreeNode root) {dfs(root);}public void dfs(TreeNode root){if(root == null){return;}TreeNode temp = root.right;root.right = root.left;root.left = temp;dfs(root.left);dfs(root.right);}}

6. 对称的二叉树

原题链接

错解：通过根节点比较子节点

没写完，太复杂

/*** 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:bool isSymmetric(TreeNode* root) {if(root == NULL)return true;return dfs(root->left,root->right);}bool dfs(TreeNode* r1,TreeNode* r2){if(r1==NULL && r2==NULL)return true;if(r1!=NULL&&r2==NULL || r1==NULL&&r2!=NULL)return false;if(r1->left->val != r2->right->val || r1->right->val != r2->left->val)return false;return dfs(r1->left,r2->right) && dfs(r1->right,r2->left);}};

正解：比较当前节点的值即可

/*** 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:bool isSymmetric(TreeNode* root) {if(root == NULL)return true;return dfs(root->left,root->right);}bool dfs(TreeNode* r1,TreeNode* r2){if(r1==NULL && r2==NULL)return true;if(r1!=NULL&&r2==NULL || r1==NULL&&r2!=NULL)return false;if(r1->val != r2->val)return false;return dfs(r1->left,r2->right) && dfs(r1->right,r2->left);}};

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public boolean isSymmetric(TreeNode root) {if(root == null){return true;}return dfs(root.left,root.right);}public boolean dfs(TreeNode l,TreeNode r){if((l == null && r != null) || (l != null && r == null)){return false;}if(l == null && r == null){return true;}if(l.val != r.val){return false;}return dfs(l.left,r.right) && dfs(l.right,r.left);}}

7. 不分行从上往下打印二叉树( 层序遍历二叉树bfs )

原题链接

/*** 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:vector<int> printFromTopToBottom(TreeNode* root) {vector<int> ans;if(root==NULL)return ans;queue<TreeNode*> q;q.push(root);while(q.size()){auto t = q.front();q.pop();ans.push_back(t->val);if(t->left != NULL)q.push(t->left);if(t->right != NULL)q.push(t->right);}return ans;}
};