二叉树先序遍历中序遍历递归及非递归解法

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].

先序遍历非递归：

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
* 思路：先访问当前节点，将当前节点入栈；
*      一直访问左子树，左子树不为空则入栈；
*      左子树为空则出栈，访问右子树，根据root = root->right重复上一句。
*/
class Solution {
public:
vector<int> preorderTraversal(TreeNode* root) {
stack<TreeNode *> s;
vector<int> v;

while(root || !s.empty()){
while(root){
v.push_back(root->val);
s.push(root);
root = root->left;
}

root = s.top();
s.pop();
root = root->right;
}
return v;
}
};

先序遍历递归求解：
/**
* 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> vec;
preorderTraversal(root,vec);

return vec;
}

void preorderTraversal(TreeNode* root,vector<int> &vec){
if(root){
vec.push_back(root->val);
preorderTraversal(root->left,vec);
preorderTraversal(root->right,vec);
}

}
};

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

For example:

Given binary tree [1,null,2,3],

1

\

/

3

return [1,3,2].

中序遍历非递归：
/**
* 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:
//当前节点不为空，入栈；
//左子树不为空入栈，左子树为空出栈，并访问右子树。重复。
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode *> s;
vector<int> v;
while(!s.empty() || root){
while(root){
s.push(root);
root=root->left;
}

root = s.top();
v.push_back(root->val);
s.pop();
root = root->right;
}
return v;
}
};

中序遍历递归：
/**
* 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> inorderTraversal(TreeNode* root) {
vector<int> vec;
inorderTraversal(root,vec);
return vec;
}
void inorderTraversal(TreeNode* root,vector<int> &vec){
if(root){
inorderTraversal(root->left,vec);
vec.push_back(root->val);
inorderTraversal(root->right,vec);
}
}
};