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[Tree]Binary Tree Preorder Traversal

时间:2015-12-12 01:39:55      阅读:146      评论:0      收藏:0      [点我收藏+]

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Total Accepted: 97599 Total Submissions: 257736 Difficulty: Medium

 

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?

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

[Tree]Binary Tree Preorder Traversal

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原文地址:http://www.cnblogs.com/zengzy/p/5040637.html

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