94. Binary Tree Inorder Traversal

题目：Binary Tree Inorder Traversal

原题链接：https://leetcode.com/problems/binary-tree-inorder-traversal/

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

For example:

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

1

\

2

/

3

return [1,3,2].

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

求二叉树中序遍历的结果，用递归和非递归的方式。

递归方式：按照中序遍历的顺序来递归就可以了

代码：

/**
* 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:
void inOrder(TreeNode* root, vector<int>& ans) {
if(root == NULL) return;
inOrder(root -> left, ans);
ans.push_back(root -> val);
inOrder(root -> right, ans);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ans;
inOrder(root, ans);
return ans;
}
};

非递归方式：用栈（stack），中序遍历的输出依次是左子树、根节点、右子树，从根节点开始一路往左子树遍历下去，并把途中所有经过的节点入栈，然后弹出一个节点，这个节点就是当前的最左节点，输出它的值，然后把它的右子树按照一路遍历左子树的方式再入栈，一直到栈空。

代码如下：

/**
* 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> ans;
stack<TreeNode*> st;
while(root != NULL) {
st.push(root);
root = root -> left;
}
while(!st.empty()) {
TreeNode* top = st.top();
st.pop();
ans.push_back(top->val);
top = top -> right;
while(top) {
st.push(top);
top = top -> left;
}
}
return ans;
}
};