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Leetcode[244]-Binary Tree Preorder Traversal

时间:2015-06-13 11:26:39      阅读:223      评论:0      收藏:0      [点我收藏+]

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].


递归遍历:

/**C++
 * 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> nums;
    vector<int> preorderTraversal(TreeNode* root) {
        if(root == NULL) return nums;
        preOrder(root);
        return nums;
    }
    void preOrder(TreeNode * root){
        if(root == NULL) return;
        nums.push_back(root->val);
        if(root->left)  preOrder(root->left);
        if(root->right) preOrder(root->right);
    }   
};

非递归遍历C++:

vector<int> preorderTraversal(TreeNode* root) {
    vector<int > nums;
    stack<TreeNode* > stk;
    if(root == NULL) return nums;
    TreeNode * p = root;
    stk.push(p);

    while(!stk.empty()){
        p = stk.top();
        nums.push_back(p->val);
        stk.pop();
        if(p->right) stk.push(p->right);
        if(p->left) stk.push(p->left);
    }
}

Leetcode[244]-Binary Tree Preorder Traversal

原文:http://blog.csdn.net/dream_angel_z/article/details/46480593

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