二叉树遍历（递归+迭代）

前序遍历

递归版

class Solution {
public:
    void preorder(TreeNode *root, vector &res) {
        if (root == nullptr) {
            return;
        }
        res.push_back(root->val);
        preorder(root->left, res);
        preorder(root->right, res);
    }

    vector preorderTraversal(TreeNode *root) {
        vector res;
        preorder(root, res);
        return res;
    }
};

迭代版

class Solution {
public:
    vector preorderTraversal(TreeNode* root) {
        vector res;
        if (root == nullptr) {
            return res;
        }

        stack stk;
        TreeNode* node = root;
        while (!stk.empty() || node != nullptr) {
            while (node != nullptr) {
                res.emplace_back(node->val);
                stk.emplace(node);
                node = node->left;
            }
            node = stk.top();
            stk.pop();
            node = node->right;
        }
        return res;
    }
};

---

中序遍历

递归版

class Solution {
public:
    void inorder(TreeNode* root, vector& res) {
        if (!root) {
            return;
        }
        inorder(root->left, res);
        res.push_back(root->val);
        inorder(root->right, res);
    }
    vector inorderTraversal(TreeNode* root) {
        vector res;
        inorder(root, res);
        return res;
    }
};

迭代版

class Solution {
public:
    vector inorderTraversal(TreeNode* root) {
        vector res;
        stack stk;
        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.push(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            res.push_back(root->val);
            root = root->right;
        }
        return res;
    }
};

---

后序遍历

递归版

class Solution {
public:
    void postorder(TreeNode *root, vector &res) {
        if (root == nullptr) {
            return;
        }
        postorder(root->left, res);
        postorder(root->right, res);
        res.push_back(root->val);
    }

    vector postorderTraversal(TreeNode *root) {
        vector res;
        postorder(root, res);
        return res;
    }
};

迭代版（这个有点难度，要记录一个prev）

class Solution {
public:
    vector postorderTraversal(TreeNode *root) {
        vector res;
        if (root == nullptr) {
            return res;
        }

        stack stk;
        TreeNode *prev = nullptr;
        while (root != nullptr || !stk.empty()) {
            while (root != nullptr) {
                stk.emplace(root);
                root = root->left;
            }
            root = stk.top();
            stk.pop();
            if (root->right == nullptr || root->right == prev) {
                res.emplace_back(root->val);
                prev = root;
                root = nullptr;
            } else {
                stk.emplace(root);
                root = root->right;
            }
        }
        return res;
    }
};

参考链接
https://leetcode-cn.com/problems/binary-tree-inorder-traversal/solution/er-cha-shu-de-zhong-xu-bian-li-by-leetcode-solutio/
https://leetcode-cn.com/problems/binary-tree-postorder-traversal/solution/er-cha-shu-de-hou-xu-bian-li-by-leetcode-solution/
https://leetcode-cn.com/problems/binary-tree-preorder-traversal/solution/er-cha-shu-de-qian-xu-bian-li-by-leetcode-solution/