94. Binary Tree Inorder Traversal

Given the root of a binary tree, return the inorder traversal of its nodes’ values.

Follow up: Recursive solution is trivial, could you do it iteratively?

• Time : O(n)
• Space : O(n)
  • also recursive solution use O(n) actual. (call stack)

/**
 * 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 {
    vector<TreeNode*> st;
    void push(TreeNode* node) {
        while(node) {
            st.push_back(node);
            node = node->left;
        }
    }
public:
    vector<int> inorderTraversal(TreeNode* root) {
        if(!root) return {};
        vector<int> res;
        push(root);
        while(!st.empty()) {
            auto pop = st.back(); st.pop_back();
            res.push_back(pop->val);
            if(pop->right)
                push(pop->right);
        }
        return res;
    }
};