[LeetCode] Binary Tree Preorder Traversal （非遞歸的先序遍歷）

[LeetCode] Binary Tree Preorder Traversal （非遞歸的先序遍歷）

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

解題思路：

二叉樹的前序遍歷。題目要求用非遞歸的方法解答。那我們先看一下遞歸方法的解法。

1、遞歸解法。

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

我們可以用兩個數據結構來存儲中間狀態。用隊列來存儲左孩子，用棧來存儲右孩子。優先遍歷所有左孩子。

/**
 * 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 preorderTraversal(TreeNode* root) {
        vector result;
        queue l;
        stack 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;
    }
};