Binary Tree Preorder Traversal (Medium)

Description

Given a binary tree, return the preorder traversal of its nodes’ values.

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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?

Analysis

非递归实现树的先序遍历。跟中序差不多。区别是存的数的时机不同。先序在插入节点的同时插入其节点的值

My Solution

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//C++
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    vector<int> ans;
    stack<TreeNode *> node;
public:
    vector<int> preorderTraversal(TreeNode *root) {
        if(root==NULL)
            return ans;
        node.push(root);
        ans.push_back(root->val);
        while(!node.empty()){
            while(root && root->left){
                node.push(root->left);
                root = root->left;
                ans.push_back(root->val);
            }
            TreeNode *t = node.top();
            node.pop();
            if(t->right){
                root = t->right;
                node.push(root);
                ans.push_back(root->val);
            }
        }
        // for(int i = 0;i<(int)ans.size();i++)
            // printf("%d,",ans[i]);
        return ans;
    }
};