Binary Tree Preorder Traversal

　　题目链接

　　题目要求：

　　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?

递归解法

　　对这道题，用递归是最简单的了，具体程序如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     vector<int> preorderTraversal(TreeNode* root) {
13         vector<int> rVec;
14         preorderTraversal(root, rVec);
15         return rVec;
16     }
17
18     void preorderTraversal(TreeNode *tree, vector<int>& rVec)
19     {
20         if(!tree)
21             return;
22
23         rVec.push_back(tree->val);
24         preorderTraversal(tree->left, rVec);
25         preorderTraversal(tree->right, rVec);
26     }
27 };

非递归解法

　　非递归解法相对就要难很多了，理解起来比较抽象。在这里我们需要借助栈。当左子树遍历完后，需要回溯并遍历右子树。具体程序如下：

 1 vector<int> preorderTraversal(TreeNode* root) {
 2     vector<int> rVec;
 3     stack<TreeNode *> st;
 4     TreeNode *tree = root;
 5     while(tree || !st.empty())
 6     {
 7         if(tree)
 8         {
 9             st.push(tree);
10             rVec.push_back(tree->val);
11             tree = tree->left;
12         }
13         else
14         {
15             tree = st.top();
16             st.pop();
17             tree = tree->right;
18         }
19     }
20
21     return rVec;
22 }

Binary Tree Inorder Traversal

　　题目链接

　　题目要求：

　　Given a binary tree, return the inorder traversal of its nodes‘ values.

　　For example:
　　Given binary tree {1,#,2,3},

   1
         2
    /
   3

　　return [1,3,2].

　　Note: Recursive solution is trivial, could you do it iteratively?

　　这道题的非递归解法跟上题的非递归解法基本一致，只是访问的节点时机不一样，具体程序如下：

 1 vector<int> inorderTraversal(TreeNode* root) {
 2     vector<int> rVec;
 3     stack<TreeNode *> st;
 4     TreeNode *tree = root;
 5     while(tree || !st.empty())
 6     {
 7         if(tree)
 8         {
 9             st.push(tree);
10             tree = tree->left;
11         }
12         else
13         {
14             tree = st.top();
15             rVec.push_back(tree->val);
16             st.pop();
17             tree = tree->right;
18         }
19     }
20
21     return rVec;
22 }

Binary Tree Postorder Traversal

　　题目链接

　　题目要求：

　　Given a binary tree, return the postorder traversal of its nodes‘ values.

　　For example:
　　Given binary tree {1,#,2,3},

   1
         2
    /
   3

　　return [3,2,1].

　　Note: Recursive solution is trivial, could you do it iteratively?