栈和递归的关系 144:Binary Tree Preorder Traversal

前序遍历：根左右

//用栈来实现非递归解法
/*** 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> preorderTraversal(TreeNode* root) {vector<int> res;if(root == NULL)return res;stack<TreeNode*> stack;stack.push(root);while(!stack.empty()){TreeNode* c = stack.top();stack.pop();res.push_back(c->val);if(c->right)stack.push(c->right);if(c->left)stack.push(c->left);}return res;}
};

//递归解法，注意res是全局的，要放在遍历函数外面
/*** 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> res;vector<int> preorderTraversal(TreeNode* root) {if(root){res.push_back(root->val);preorderTraversal(root->left);preorderTraversal(root->right);}return res;}
};

中序遍历：左根右

/*** 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> inorderTraversal(TreeNode* root) {stack<TreeNode*> st;vector<int> res;if(root == NULL) return res;  //若根节点为空则返回空TreeNode* p = root;while(p || !st.empty()){while(p){//先将根节点入栈，再将它所有的左节点入栈
                st.push(p);p = p->left;   }p = st.top();st.pop();res.push_back(p->val);p = p->right;}return res;}
};

 后序遍历：左右根

可以使其遍历顺序为根左右，然后逆序插入vector中，即每次在vector的头部插入结点值。在压入栈时先压入右结点再压入左结点则在出栈时就是先左后右了。

//解法一
/*** 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> postorderTraversal(TreeNode* root) {if(!root) return {};vector<int> res;stack<TreeNode*> s{{root}};while(!s.empty()){TreeNode* t = s.top();s.pop();res.insert(res.begin(), t->val);if(t->left) s.push(t->left);if(t->right) s.push(t->right);}return res;}
};

解法二：关键是判断当前这个结点：

1）它如果有左右结点是否已经入栈，若没有入栈则先将它的右结点入栈，再左结点入栈；如果它的左右结点已经入栈，则这个结点就可以直接加入容器vector里了。

2）如果它是叶子结点（没有左右结点），则直接加入vector中。

/*** 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> postorderTraversal(TreeNode* root) {if(!root) return {};vector<int> res;stack<TreeNode*> s{{root}};TreeNode* head = root;   //head初始化while(!s.empty()){TreeNode* t = s.top();if((!t->left && !t->right) || t->left==head || t->right==head){//当t为叶子结点 或t的左结点或右结点为head，即已经入栈了res.push_back(t->val);s.pop();head = t;}else{if(t->right) s.push(t->right);if(t->left) s.push(t->left);}}return res;}
};

 

leetcode已经分别定义了类NestedInteger中的三个函数：

1）isInteger()：若当前这个NestedInteger是单个整数返回true；

2）getInteger()：返回当前这个单个的整数；

3）getList()：返回当前这个列表。

/*** // This is the interface that allows for creating nested lists.* // You should not implement it, or speculate about its implementation* class NestedInteger {*   public:*     // Return true if this NestedInteger holds a single integer, rather than a nested list.*     bool isInteger() const;**     // Return the single integer that this NestedInteger holds, if it holds a single integer*     // The result is undefined if this NestedInteger holds a nested list*     int getInteger() const;**     // Return the nested list that this NestedInteger holds, if it holds a nested list*     // The result is undefined if this NestedInteger holds a single integer*     const vector<NestedInteger> &getList() const;* };*/
class NestedIterator {
public:stack<NestedInteger> s;NestedIterator(vector<NestedInteger> &nestedList) {for(int i=nestedList.size()-1; i>=0; i--)//倒序插入 是为了弹出时是正序的
            s.push(nestedList[i]);}int next() {//返回下一项的值NestedInteger t = s.top();s.pop();return t.getInteger();  //获取对应整数
        }bool hasNext() {//若下一项有值，返回truewhile(!s.empty()){NestedInteger t = s.top();if(t.isInteger())return true;   //若下一个数是单个整数，返回true
            s.pop();//若下一个数是一个列表,使用getList()得到列表的每个整数倒序插入到栈中for(int i=t.getList().size()-1; i>=0; i--)s.push(t.getList()[i]);}return false;}
};/*** Your NestedIterator object will be instantiated and called as such:* NestedIterator i(nestedList);* while (i.hasNext()) cout << i.next();*/