[LeetCode] Binary Tree Preorder Traversal （非递归的先序遍历）

Binary Tree Preorder Traversal

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

解题思路：

二叉树的前序遍历。题目要求用非递归的方法解答。那我们先看一下递归方法的解法。

1、递归解法。

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        preorderHelper(result, root);
        return result;
    }

    void preorderHelper(vector<int>& result, TreeNode* root){
        if(root==NULL){
            return;
        }
        result.push_back(root->val);
        preorderHelper(result, root->left);
        preorderHelper(result, root->right);
    }
};

2、非递归解法

我们可以用两个数据结构来存储中间状态。用队列来存储左孩子，用栈来存储右孩子。优先遍历所有左孩子。

/**
 * 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> preorderTraversal(TreeNode* root) {
        vector<int> result;
        queue<TreeNode*> l;
        stack<TreeNode*> r;
        if(root!=NULL){
            l.push(root);
        }
        while(!l.empty()||!r.empty()){
            TreeNode* node = NULL;
            if(!l.empty()){
                node=l.front();
                l.pop();
            }else{
                node=r.top();
                r.pop();
            }
            result.push_back(node->val);
            if(node->left!=NULL){
                l.push(node->left);
            }
            if(node->right!=NULL){
                r.push(node->right);
            }
        }

        return result;
    }
};