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.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

分析：这道题用递归方法很好做，但是此题却提示用循环方法解决。首先我用递归方法，把层次遍历转化为中序遍历，看代码实现就会明白：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void traversal(TreeNode* root,vector<int> &data)
    {
        if(root==NULL)
            return;
        traversal(root->left,data);
        data.push_back(root->val);
        traversal(root->right,data);
    }
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> data;
        traversal(root,data);
        return data;
    }
};

循环实现，这题就有点难度了，不过没关系，我们可以借用栈空间来解决这题。树中某结点不为空，则将其左结点压入栈中，如果没有，则弹出该节点，将该节点的值存入vector中，再压入出栈元素的右结点，依次进行循环遍历，直到栈为空为止。

/**
 * 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) {
        stack<TreeNode*> StackTree;
        vector<int> data;
        if(root==NULL)
            return data;
        TreeNode *CurrentNode=root;
        while(!StackTree.empty()||CurrentNode)
        {
            if(CurrentNode!=NULL)
            {
                StackTree.push(CurrentNode);
                CurrentNode=CurrentNode->left;
            }
            else
            {
                CurrentNode=StackTree.top();
                StackTree.pop();
                data.push_back(CurrentNode->val);
                CurrentNode=CurrentNode->right;
            }
        }
        return data;
    }
};