反转二叉树
递归写法
很简单
class Solution {
public:TreeNode* invertTree(TreeNode* root) {if(root==nullptr)return root;TreeNode* tmp;tmp=root->left;root->left=root->right;root->right=tmp;invertTree(root->left);invertTree(root->right);return root;}
};
迭代写法
先序遍历
class Solution {
public:TreeNode* invertTree(TreeNode* root) {stack<TreeNode*> sta;if(root)sta.push(root);TreeNode* cur;while(!sta.empty()){cur=sta.top();sta.pop();if(cur->left)sta.push(cur->left);if(cur->right)sta.push(cur->right);TreeNode* tmp=cur->left;cur->left=cur->right;cur->right=tmp;}return root;}
};
后序也可以；
但是中序遍历不行；
class Solution {
public:TreeNode* invertTree(TreeNode* root) {stack<TreeNode*> sta;if(root)sta.push(root);TreeNode* cur;while(!sta.empty()){cur=sta.top();sta.pop();if(cur){if(cur->right)sta.push(cur->right);sta.push(cur);sta.push(nullptr);if(cur->left)sta.push(cur->left);}else{cur=sta.top();sta.pop();TreeNode* tmp=cur->left;cur->left=cur->right;cur->right=tmp;}}return root;}
};
但是这个是可以过oj的
对称二叉树
反转之后再和原来的树比较，一模一样就是对称的。
但是这样还要复制一棵树，浪费空间。
应该两个指针同时遍历，镜像遍历：
只能是后序遍历？
- 前序遍历
这种情况会出现假阳性，根本在于前序遍历不能唯一标识一棵树
- 中序遍历
同理，中序遍历也不能唯一标识一棵树：加法的二义性
- 后序遍历
后序遍历也不能唯一标识一棵树，例子同前序遍历的例子。
所以：并非只有后序遍历才行
class Solution {
public:bool Compare(TreeNode* cur1,TreeNode* cur2){if(cur1->left&&cur2->right){if(cur1->left->val!=cur2->right->val)return 0;}else if(cur1->left&&!cur2->right){return 0;}else if(!cur1->left&&cur2->right){return 0;}if(cur2->left&&cur1->right){if(cur2->left->val!=cur1->right->val)return 0;}else if(cur2->left&&!cur1->right){return 0;}else if(!cur2->left&&cur1->right){return 0;}return 1;}bool isSymmetric(TreeNode* root) {stack<TreeNode*> sta1;stack<TreeNode*> sta2;if(root){sta1.push(root);sta2.push(root);}TreeNode* cur1,*cur2;while(!sta1.empty()&&!sta2.empty()){cur1=sta1.top();sta1.pop();cur2=sta2.top();sta2.pop();if(!Compare(cur1,cur2))return 0;if(cur1->right)sta1.push(cur1->right);if(cur1->left)sta1.push(cur1->left);if(cur2->left)sta2.push(cur2->left);if(cur2->right)sta2.push(cur2->right);}return 1;}
};
只需要在比较时添加上子节点的因素，就可以避免假阳性
二叉树最大深度
层序遍历
class Solution {
public:int maxDepth(TreeNode* root) {queue<TreeNode*> que;if(root) que.push(root);int cur_size=1,nxt_size=0;int depth=0;while(!que.empty()){TreeNode* cur=que.front();que.pop();if(cur->left){nxt_size++;que.push(cur->left);}if(cur->right){nxt_size++;que.push(cur->right);}if(--cur_size==0){depth++;cur_size=nxt_size;nxt_size=0;}}return depth;}
};
更简洁的写法：
class Solution {
public:int maxDepth(TreeNode* root) {if (root == NULL) return 0;int depth = 0;queue<TreeNode*> que;que.push(root);while(!que.empty()) {int size = que.size();depth++; // 记录深度for (int i = 0; i < size; i++) {TreeNode* node = que.front();que.pop();if (node->left) que.push(node->left);if (node->right) que.push(node->right);}}return depth;}
};
二叉树最小深度
层序遍历
class Solution {
public:int minDepth(TreeNode* root) {queue<TreeNode*> que;if(root) que.push(root);int cur_size=1,nxt_size=0;int depth=0;while(!que.empty()){TreeNode* cur=que.front();que.pop();if(!cur->left&&!cur->right)return depth+1;if(cur->left){nxt_size++;que.push(cur->left);}if(cur->right){nxt_size++;que.push(cur->right);}if(--cur_size==0){depth++;cur_size=nxt_size;nxt_size=0;}}return depth;}
};
->right){
nxt_size++;
que.push(cur->right);
}
if(–cur_size==0){
depth++;
cur_size=nxt_size;
nxt_size=0;
}
}return depth;
}
};
> 判断有没有叶节点
