Binary Tree Inorder Traversal——二叉树的中序遍历

原题：

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

=>给定一个二叉树，返回所有节点的中序遍历

For example:

=>例如
 Given binary tree {1,#,2,3},

=>给定二叉树如下：

   1
    \
     2
    /
   3

return [1,3,2].

=>返回[1,3,2]

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

=>递归的算法很简单，能否不递归实现？

/**
 * Definition for binary tree
 * 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) {
        
    }
};

晓东解析：

    这个题目和之前的先序遍历以及后序遍历是一样的，就不多做解释了，见相应的博文：http://blog.csdn.net/u011960402/article/details/14517135以及http://blog.csdn.net/u011960402/article/details/15499903。

代码实现：

    1）递归实现：

/**
 * Definition for binary tree
 * 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;
        vector<int> left;
        vector<int> right;
        
        if(NULL == root) return result;
        
        left = inorderTraversal(root->left);
        if(left.size() != 0)
            result.insert(result.end(), left.begin(), left.end());
        result.push_back(root->val);
        right = inorderTraversal(root->right);
        if(right.size() != 0)
            result.insert(result.end(), right.begin(), right.end());
        
        return result;
        
    }
};

执行结果：

67 / 67 test cases passed.	Status: Accepted
Runtime: 36 ms

  2）非递归实现

/**
 * Definition for binary tree
 * 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*> TreeStack;
        
        if(NULL == root) return result;
        
        while(root || !TreeStack.empty()){
            while(root){
                TreeStack.push(root);
                root =  root->left;
            }
            root = TreeStack.top();
            result.push_back(root->val);
            TreeStack.pop();
            root =  root->right;
        }
        
    }
};

执行结果：

67 / 67 test cases passed.	Status: Accepted
Runtime: 36 ms

若您有更好的算法，欢迎提出指正。