LeetCode 94. 二叉树的中序遍历（）

原题解

题目

约束

题解

方法一

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

方法二

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> 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;
    }
};

方法三

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        TreeNode *predecessor = nullptr;

        while (root != nullptr) {
            if (root->left != nullptr) {
                // predecessor 节点就是当前 root 节点向左走一步，然后一直向右走至无法走为止
                predecessor = root->left;
                while (predecessor->right != nullptr && predecessor->right != root) {
                    predecessor = predecessor->right;
                }
                
                // 让 predecessor 的右指针指向 root，继续遍历左子树
                if (predecessor->right == nullptr) {
                    predecessor->right = root;
                    root = root->left;
                }
                // 说明左子树已经访问完了，我们需要断开链接
                else {
                    res.push_back(root->val);
                    predecessor->right = nullptr;
                    root = root->right;
                }
            }
            // 如果没有左孩子，则直接访问右孩子
            else {
                res.push_back(root->val);
                root = root->right;
            }
        }
        return res;
    }
};