LeetCode 94. Binary Tree Inorder Traversal

Solution1：递归版
二叉树的中序遍历递归版是很简单的，中序遍历的迭代版需要特殊记一下！
迭代版链接：https://blog.csdn.net/allenlzcoder/article/details/79837841

/*** 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) {vector<int> res;my_Inordertraversal(root, res);return res;}void my_Inordertraversal(TreeNode* root, vector<int>& res) {if (!root) return;my_Inordertraversal(root->left, res);res.push_back(root->val);my_Inordertraversal(root->right, res);}
};

Solution2：迭代版
对于任一当前结点P：
1)若P的左孩子不为空，则将P入栈并将P的左孩子置为当前节点，然后对当前结点再进行重复处理1)；
2)若P的左孩子为空，则取栈顶元素s.top()并进行出栈操作，访问s.top()->val，然后将当前结点置为的s.top()的右孩子；若s.top()->right == NULL 即当前结点为NULL，重复处理2)；
3)直到P为NULL并且栈为空则遍历结束。
【关键】：若当前结点为空，则进行取栈顶元素操作（访问结点元素）并把当前结点设置为栈顶结点的右孩子。

/*** 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) {vector<int> res;if (!root) return res;TreeNode* cur = root;stack<TreeNode*> my_stack;while(!my_stack.empty() || cur != NULL) {if(cur == NULL) {TreeNode* temp = my_stack.top();my_stack.pop();res.push_back(temp->val);cur = temp->right;} else {my_stack.push(cur);cur = cur->left;}}return res;}
};

稍微修改后的迭代版

/*** 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) {if (!root) return {};vector<int> res;stack<TreeNode* > s_tree;s_tree.push(root);TreeNode *cur = root->left;while (!s_tree.empty()) {while (cur) { //把最左边的结点压入栈中s_tree.push(cur);cur = cur->left;}//访问当前结点的值TreeNode *temp = s_tree.top(); res.push_back(temp->val);    s_tree.pop();    cur = temp->right; if (cur) {s_tree.push(cur);cur = cur->left;}}return res;}
};

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