94. 二叉树的中序遍历

题目链接：leetcode.

递归

/**
 * 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 {
    void dfs(TreeNode* root, vector<int>& ans)
    {
        if(root == nullptr)
        {
            return;
        }
        dfs(root -> left, ans);
        ans.emplace_back(root -> val);
        dfs(root -> right, ans);
    }
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        dfs(root, ans);
        return ans;
    }
};

我写的迭代，（用哈希表存储是否访问过，low

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        if(root == nullptr)
        {
            return {};
        }
        vector<int> ans;
        stack<TreeNode*> S;
        unordered_set<TreeNode*> hash;
        S.push(root);
        while(!S.empty())
        {
            TreeNode* tmp = S.top();
            while(tmp -> left && hash.count(tmp -> left) == 0)
            {
                S.push(tmp -> left);
                tmp = S.top();
            }
            S.pop();
            ans.emplace_back(tmp -> val);
            hash.insert(tmp);
            if(tmp -> right)
                S.push(tmp -> right);
        }
        return ans;
    }
};

官解的迭代

/*
执行用时：4 ms, 在所有 C++ 提交中击败了41.18%的用户
内存消耗：8.1 MB, 在所有 C++ 提交中击败了77.51%的用户
*/
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> ans;
        stack<TreeNode*> S;
        TreeNode* p = root;
        while(!S.empty() || p != nullptr)
        {
            while(p != nullptr)
            {
            	S.push(p);
            	p = p -> left;
            }
            p = S.top();
            S.pop();
            ans.emplace_back(p -> val);
            p = p -> right;
        }
        return ans;
    }
};