leetcode: Binary Tree Preorder Traversal

标签：

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].

Note: Recursive solution is trivial, could you do it iteratively?

题目描述：

这是一个二叉树的先序遍历的题目，思路是直接用栈去实现。具体代码是借鉴的别人的，没怎么看懂，先把代码附上，以后慢慢体会。

代码实现：

/**
 * 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>re;
        const TreeNode *p;
        stack<const TreeNode *>s;
        p=root;
        if(p!=NULL)
        {
            s.push(p);
        }
        while(!s.empty())
        {
            p=s.top();
            s.pop();
            re.push_back(p->val);
            if(p->right!=NULL)
            {
                s.push(p->right);
            }
            if(p->left!=NULL)
            {
                s.push(p->left);
            }
        }
        return re;
    }
};

leetcode: Binary Tree Preorder Traversal

标签：

原文地址：http://blog.csdn.net/flyljg/article/details/48006391