力扣144. 二叉树的前序遍历145. 二叉树的后序遍历94. 二叉树的中序遍历（非递归版）

前序遍历

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> st;
        st.push(root);
        while(!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            if(node != nullptr) result.push_back(node->val);
            else continue;
            st.push(node->right);
            st.push(node->left);
        }

        return result;
    }
};

后序遍历
这里的思路是交换了

			st.push(node->right);
            st.push(node->left);

的顺序，从先序列遍历的根左右变成了根右左，接着通过reverse变成了左右根，即后序遍历。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> st;
        st.push(root);
        while(!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            if(node != nullptr) result.push_back(node->val);
            else continue;
            st.push(node->left);
            st.push(node->right);
        }
        reverse(result.begin(), result.end());

        return result;
    }
};

中序遍历
写法一：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> st;
        TreeNode* cur = root;
        while (cur != nullptr || !st.empty()) {
            if(cur != nullptr) {
                st.push(cur);
                cur = cur->left;
            } else {
                cur = st.top();
                st.pop();
                result.push_back(cur->val);
                cur = cur->right;
            }
        }

        return result;
    }
};

写法二：

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode*> st;
        TreeNode* node = root;

        while (node || !st.empty()) {  // 遍历整个树
            while (node) {             // 先遍历到最左
                st.push(node);
                node = node->left;
            }
            node = st.top();
            st.pop();
            result.push_back(node->val);
            node = node->right;  // 进入右子树
        }
        
        return result;
    }
};