[LeetCode]Binary Tree Inorder Traversal

题目描述:(链接)

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,3,2].

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

解题思路：

迭代版:

/**
 * 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> inorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode *> cache;
        TreeNode *p = root;
        
        while (!cache.empty() || p != nullptr) {
            if (p != nullptr) {
                cache.push(p);
                p = p->left;
            } else {
                p = cache.top();
                cache.pop();
                result.push_back(p->val);
                p = p->right;
            }
        }
        
        return result;
    }
};

 递归版：

/**
 * 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> inorderTraversal(TreeNode* root) {
        if (root != nullptr) {
            inorderTraversal(root->left);
            result.push_back(root->val);
            inorderTraversal(root->right);
        }
        
        return result;
    }
private:
    vector<int> result;
};