66.二叉树的前序遍历

描述

给出一棵二叉树，返回其节点值的前序遍历。

您在真实的面试中是否遇到过这个题？  是

样例

给出一棵二叉树 {1,#,2,3},

   1
    \
     2
    /
   3

 返回 [1,2,3].

挑战

你能使用非递归实现么？

遍历二叉树使用递归十分简单：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
    vector<int> m_vector;
     
    vector<int> preorderTraversal(TreeNode * root) {
        // write your code here
        if(root!=NULL)
            search(root);
        return m_vector;
    }
    
    void search(TreeNode *m_node){
        m_vector.push_back(m_node->val);
        if(m_node->left!=NULL)
            search(m_node->left);
        if(m_node->right!=NULL)
            search(m_node->right);
    }
};

采用非递归的方式，可以使用栈来辅助操作：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
     
    vector<int> preorderTraversal(TreeNode * root) {
        // write your code here
        stack<TreeNode*> m_stack;
        vector<int> m_vector;
        TreeNode * m_node=root;
        if(root!=NULL)
            m_stack.push(m_node);
        while(!m_stack.empty()){
            m_vector.push_back(m_node->val);
            if(m_node->right!=NULL){
                m_stack.push(m_node->right);
            }
            if(m_node->left!=NULL){
                m_node=m_node->left;
            }else{
                m_node=m_stack.top();
                m_stack.pop();
            }
        }
        return m_vector;
    }
};

但我第一次犯了错误思考了很久：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: A Tree
     * @return: Preorder in ArrayList which contains node values.
     */
     
    vector<int> preorderTraversal(TreeNode * root) {
        // write your code here
        stack<TreeNode> m_stack;
        vector<int> m_vector;
        TreeNode * m_node=root;
        if(root!=NULL)
            m_stack.push(*m_node);
        while(!m_stack.empty()){
            m_vector.push_back(m_node->val);
            if(m_node->right!=NULL){
                m_stack.push(*m_node->right);
            }
            if(m_node->left!=NULL){
                m_node=m_node->left;
            }else{
                m_node=&m_stack.top();
                m_stack.pop();
            }
        }
        return m_vector;
    }
};

至今不能明白为什么

m_stack.push

之后m_code的值会发生改变，希望有明白的同学可以不吝赐教。