4.35 leetcode -35 binary-tree-inorder-traversal

题目描述

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}".

所以，需要非递归方法的前序遍历是么。

还好写的出来

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