【树】94. 二叉树的中序遍历

题目：

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？

 

解法：

方法一：递归实现

 

/**
 * 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;
        if (NULL == root)
        {
            return result;
        }

        vector<int> left_result = inorderTraversal(root->left);
        vector<int> right_result = inorderTraversal(root->right);

        // 注意下面, 需要考虑下
        for (int i = 0; i < left_result.size(); i++)
        {
            result.push_back(left_result[i]);
        }
        result.push_back(root->val);
        for (int i = 0; i < right_result.size(); i++)
        {
            result.push_back(right_result[i]);
        }

        return result;
    }
};

 

 

 

 

方法二：迭代算法

 

/**
 * 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> ret;
        if (NULL ==  root)
        {
            return ret;
        }

        stack<TreeNode *> trace;
        TreeNode *tmp = root;

        while (tmp != NULL || trace.empty() != true)
        {
            while (tmp != NULL)
            {
                trace.push(tmp);
                tmp = tmp->left;
            }

            if (trace.empty() != true)
            {
                tmp = trace.top();
                trace.pop();
                ret.push_back(tmp->val);
                tmp = tmp->right;
            }
        }

        return ret;
    }
};