LeetCode:Binary Tree Preorder Traversal(二叉树的先序遍历)

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

解法一：用栈实现（递归本质）

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        stack<TreeNode *> node_stack;
        if(root!=NULL)
            node_stack.push(root);
        TreeNode *node;
        
        while(!node_stack.empty())
        {   
            node=node_stack.top();
            result.push_back(node->val);
            node_stack.pop();
            
            if(node->right!=NULL) node_stack.push(node->right);
            if(node->left!=NULL) node_stack.push(node->left);
        }
        
        return result;       
        
    }
};

 解法二：递归方式

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        preorder(root,result);
        return result;
        
        
    }
    void preorder(TreeNode *node,vector<int> &result)
    {   
        if(node!=NULL)
        {
            result.push_back(node->val);
            preorder(node->left,result);
            preorder(node->right,result);
        }
        
    }
};

 解法三：morris遍历方式 有待更新。