[LeetCode]Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

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?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

 

中序遍历。左-根-右。迭代使用stack。

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

递归的方法很统一。先序，中序，后续，只要修改遍历的顺序即可。

/**
 * 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;
        traversal(root,result);
        return result;
    }
    void traversal(TreeNode* root,vector<int>& ret)
    {
        if(root)
        {
            traversal(root->left,ret);
            ret.push_back(root->val);
            traversal(root->right,ret);
        }
    }
};