144. Binary Tree Preorder Traversal &&94. Binary Tree Inorder Traversal

题目：先序遍历

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?

---

翻译

用先序遍历排列：

源码

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

---

题目：中序遍历

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

For example:Given binary tree {1,#,2,3},   1    \     2    /   3return [1,3,2].

给定一个二叉树，返回节点值得中序遍历。
中序遍历：左中右；

/** * 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> result;    vector<int> inorderTraversal(TreeNode* root) {        if(root!=NULL){            inorderTraversal(root->left);            result.push_back(root->val);            inorderTraversal(root->right);        }        return result;    }};

---

拓展：

先序遍历二叉树：（若二叉树为空，则空操作）
先访问根节点；先序遍历左子树，先序遍历右子树；

中序遍历二叉树：
中序遍历左子树；访问根节点；中序遍历右子树；

后序遍历二叉树：
后序遍历左子树；后序遍历右子树；访问根节点。