[LeetCode] Binary Tree Inorder Traversal

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [1,3,2].

递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorder;
    vector<int> inorderTraversal(TreeNode* root) {
        if (root == nullptr)
            return inorder;
        inorderTraversal(root->left);
        inorder.push_back(root->val);
        inorderTraversal(root->right);
        return inorder;
    }
};
// 0 ms

迭代

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorder;
    vector<int> inorderTraversal(TreeNode* root) {
        if (root == nullptr)
            return inorder;
        stack<TreeNode*> stk;
        while (root != nullptr || !stk.empty()) {
            if (root != nullptr) {
                stk.push(root);
                root = root->left;
            }
            else {
                root = stk.top();
                stk.pop();
                inorder.push_back(root->val);
                root = root->right;
            }
        }
        return inorder;
    }
};
// 3 ms