binary-tree-preorder-traversal 递归和非递归前序遍历二叉树

题目描述

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?

递归

/**
 * 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> ret;
    vector<int> preorderTraversal(TreeNode *root) {
        if(!root) return ret;
        ret.push_back(root->val);
        preorderTraversal(root->left);
        preorderTraversal(root->right);
        return ret;
    }
};

非递归

/**
 * 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) {
        vector<int> ret;
        stack<TreeNode *> st;
        TreeNode* cur;
        if(!root) return ret;
        st.push(root);
        while(!st.empty()){
            cur = st.top(); st.pop();
            ret.push_back(cur->val);
            if(cur->right) st.push(cur->right);
            if(cur->left) st.push(cur->left);
        }
        return ret;
    }
};