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[LeetCode] Binary Tree Preorder Traversal

时间:2017-09-10 12:09:22      阅读:188      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

/**
 * 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> preorder;
    vector<int> preorderTraversal(TreeNode* root) {
        if (root == nullptr)
            return preorder;
        preorder.push_back(root->val);
        preorderTraversal(root->left);
        preorderTraversal(root->right);
        return preorder;
    }
};
// 0 ms

迭代

/**
 * 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> preorder;
    vector<int> preorderTraversal(TreeNode* root) {
        if (root == nullptr)
            return preorder;
        stack<TreeNode*> stk;
        while (root != nullptr || !stk.empty()) {
            if (root != nullptr) {
                stk.push(root);
                preorder.push_back(root->val);
                root = root->left;
            }
            else {
                root = stk.top();
                stk.pop();
                root = root->right;
            }
        }
        return preorder;
    }
};
// 0 ms

 

[LeetCode] Binary Tree Preorder Traversal

原文:http://www.cnblogs.com/immjc/p/7500534.html

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