Binary Tree Preorder Traversal

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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 binary tree
 * 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) {
        stack<TreeNode*> nodestack;
        vector<int> arr(0);
        TreeNode *node;
        if(root == NULL) return arr;
        nodestack.push(root);
        while(!nodestack.empty()){
            node=nodestack.top();
            arr.push_back(node->val);
            nodestack.pop();
            if(node->right!=NULL) nodestack.push(node->right);
            if(node->left!=NULL) nodestack.push(node->left);
        }
        return arr;
    }
};

Binary Tree Preorder Traversal

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原文地址：http://blog.csdn.net/uj_mosquito/article/details/42024869