力扣算法题—094中序遍历二叉树【】

给定一个二叉树，返回它的中序 遍历。

示例:

输入: [1,null,2,3]
   1
    
     2
    /
   3

输出: [1,3,2]

进阶: 递归算法很简单，你可以通过迭代算法完成吗？

//递归遍历
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int>res;
        helper(res, root);
        return res;
    }
    void helper(vector<int>&res, TreeNode* root) {
        if (root==NULL)
            return;
        if (root->left)
            helper(res, root->left);
        res.push_back(root->val);
        if (root->right)
            helper(res, root->right);
    }
};


//非递归遍历
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int>res;
        stack<TreeNode*>s;
        TreeNode* p = root;
        while (p || !s.empty()) {
            while (p) {
                s.push(p);
                p = p->left;
            }//找到最左节点
            p = s.top();
            s.pop();
            res.push_back(p->val);
            p = p->right;
        }
        return res;
    }
};


//另一种解法
// Non-recursion and no stack
class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        if (!root) return res;
        TreeNode *cur, *pre;
        cur = root;
        while (cur) {
            if (!cur->left) {
                res.push_back(cur->val);
                cur = cur->right;
            }
            else {
                pre = cur->left;
                while (pre->right && pre->right != cur) pre = pre->right;
                if (!pre->right) {
                    pre->right = cur;
                    cur = cur->left;
                }
                else {
                    pre->right = NULL;
                    res.push_back(cur->val);
                    cur = cur->right;
                }
            }
        }
        return res;
    }
};