[leetcode] 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?

给定一棵二叉树,使用非递归的方法进行先序遍历。

用一个栈来保存回溯的节点,对于每个节点,都先访问栈顶元素,右儿子非空入栈,左儿子非空入栈的顺序进行,直到栈为空。

代码如下:

 1 class Solution {
 2     public:
 3         vector<int> preorderTraversal(TreeNode *root) {
 4             vector<int> res;
 5             stack<TreeNode *> st;
 6             TreeNode * p = root;
 7             st.push(p);
 8             if( root == NULL )
 9             {
10                 return res; 
11             }
12             while( st.size() != 0 )
13             {
14                 p = st.top();
15                 st.pop();
16                 res.push_back(p->val);
17                 if( p->right != NULL )
18                 {
19                     st.push(p->right);
20                 }
21                 if( p->left != NULL )
22                 {
23                     st.push(p->left);
24                 }
25             }
26             return res;
27         }
28 };
原文地址:https://www.cnblogs.com/jostree/p/3720904.html