网站建设广金手指排名,2017如何免费制作网站,小程序可以自己开发吗,wish跨境电商平台官网此篇皆为leetcode、牛客中的简单题型和二叉树基础操作#xff0c;无需做过多讲解#xff0c;仅付最优解。有需要的小伙伴直接私信我~
目录
1.二叉树的节点个数
2.二叉树叶子节点个数
3.二叉树第K层节点个数
4.查找值为X的节点
5.leetcode——二叉树的最大深度
6.leetc…
此篇皆为leetcode、牛客中的简单题型和二叉树基础操作无需做过多讲解仅付最优解。有需要的小伙伴直接私信我~
目录
1.二叉树的节点个数
2.二叉树叶子节点个数
3.二叉树第K层节点个数
4.查找值为X的节点
5.leetcode——二叉树的最大深度
6.leetcode——单值二叉树
7.leetcode——相同的树
8.二叉树的前序遍历
9.二叉树的中序遍历
10.二叉树的后序遍历
11.二叉树的层序遍历
12.leetcode——另一棵树的子树
13.二叉树的构建及遍历
14.leetcode——对称二叉树 1.二叉树的节点个数
int BinaryTreeSize(BTNode* root)
{return root NULL ? 0 : BinaryTreeSize(root-left) BinaryTreeSize(root-right) 1;
}
2.二叉树叶子节点个数
int BinaryTreeLeafSize(BTNode* root)
{if (root NULL)return 0;if (root-left NULL root-right NULL)return 1;return BinaryTreeLeafSize(root-left) BinaryTreeLeafSize(root-right);
}
3.二叉树第K层节点个数
int BinaryTreeLevelKSize(BTNode* root, int k)
{if (root NULL)return 0;if (k 1)return 1;return BinaryTreeLevelKSize(root-left, k - 1) BinaryTreeLevelKSize(root-right, k - 1);
}4.查找值为X的节点
BTNode* BinaryTreeFind(BTNode* root, BTDataType data)
{if (root NULL)return NULL;if (root-data data)return root;BTNode* ret1 BinaryTreeFind(root-left, data);if (ret1)return ret1;BTNode* ret2 BinaryTreeFind(root-right, data);if (ret2)return ret2;return NULL;
}
5.leetcode——二叉树的最大深度
oj链接二叉树的最大深度
int maxDepth(struct TreeNode* root) {if (root NULL)return 0;int leftDepth maxDepth(root-left);int rightDepth maxDepth(root-right);return (leftDepth rightDepth ? leftDepth : rightDepth) 1;
}
6.leetcode——单值二叉树
oj链接单值二叉树
bool isUnivalTree(struct TreeNode* root)
{if (root NULL){return true;}if (root-left root-val ! root-left-val)return false;if (root-right root-val ! root-right-val)return false;return isUnivalTree(root-left) isUnivalTree(root-right);
}
7.leetcode——相同的树
oj链接相同的树
bool isSameTree(struct TreeNode* p, struct TreeNode* q) {if (p NULL q NULL)return true;if (p NULL || q NULL)return false;if (p-val ! q-val){return false;}return isSameTree(p-left, q-left) isSameTree(p-right, q-right);
}8.二叉树的前序遍历
详细讲解二叉树的前、中、后序遍历
void PrevOrder(BTNode* root)
{if (root NULL){printf(NULL );return;}printf(%d , root-data);//前序在前PrevOrder(root-left);PrevOrder(root-right);
}
9.二叉树的中序遍历
void InOrder(BTNode* root)
{if (root NULL){printf(NULL );return;}InOrder(root-left);printf(%d , root-data);//中序在中InOrder(root-right);
}
10.二叉树的后序遍历
void PostOrder(BTNode* root)
{if (root NULL){printf(NULL );return;}PostOrder(root-left);PostOrder(root-right);printf(%d , root-data);//后序在后
}
11.二叉树的层序遍历
详细讲解看完这篇我不信你不会二叉树的层序遍历
//需自己实现队列的数据结构
void BinaryTreeLevelOrder(BTNode* root)
{Queue q;QueueInit(q);if (root)QueuePush(q, root);elsereturn;while (!QueueEmpty(q)){BTNode* front QueueFront(q);printf(%d , front-data);QueuePop(q);if(front-left)QueuePush(q, front-left);if(front-right)QueuePush(q, front-right);}printf(\n);QueueDestroy(q);
}
12.leetcode——另一棵树的子树
oj链接另一棵树的子树
//判断两树是否相同复用“相同的树”解题代码
bool isSametree(struct TreeNode* p,struct TreeNode* q)
{if(pNULL qNULL)return true;if(pNULL || qNULL)return false;if(p-val ! q-val)return false;return isSametree(p-left,q-left) isSametree(p-right,q-right);
}bool isSubtree(struct TreeNode* root, struct TreeNode* subRoot){if(rootNULL)return false;if(isSametree(root,subRoot))return true;return isSubtree(root-left,subRoot) || isSubtree(root-right,subRoot);
}
13.二叉树的构建及遍历
题目链接二叉树的遍历
注意本题虽然是二叉树的遍历但考查的却是如何通过数组的内容构建一棵二叉树。
#include stdio.h
#include stdlib.hstruct TreeNode {int val;struct TreeNode *left;struct TreeNode *right;};struct TreeNode* rebuildTree(char* str,int* pi)
{if(str[*pi] #){(*pi);return NULL;}struct TreeNode* root (struct TreeNode*)malloc(sizeof(struct TreeNode));root-val str[(*pi)];root-left rebuildTree(str,pi);root-right rebuildTree(str,pi);return root;
}void TreeDestory(struct TreeNode* root)
{if(rootNULL)return;TreeDestory(root-left);TreeDestory(root-right);free(root);
}void InOrder(struct TreeNode* root)
{if (root NULL){return;}InOrder(root-left);printf(%c , root-val);InOrder(root-right);
}
int main() {char str[100];scanf(%s,str);int i0;struct TreeNode* root rebuildTree(str,i);InOrder(root);TreeDestory(root);return 0;
}
14.leetcode——对称二叉树
oj链接对称二叉树
bool _isSymmetric(struct TreeNode* p, struct TreeNode* q)
{if (p NULL q NULL)return true;if (p NULL || q NULL)return false;if (p-val ! q-val)return false;return _isSymmetric(p-left, q-right) _isSymmetric(p-right, q-left);
}
bool isSymmetric(struct TreeNode* root) {return root NULL ? 0 : _isSymmetric(root-left, root-right);
}
