给定一个二叉树,返回它的中序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [1,3,2]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
注意迭代算法的实现,仔细思考迭代算法是怎么处理以下两种情况的:
1)
1
/
2
/
3
2)
1
/
2
/
3
\
4
/**
* 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;
inorder(&vec,root);
return vec;
}
void inorder(vector<int> *vec,TreeNode* root){
if(root==NULL) return;
inorder(vec,root->left);
vec->push_back(root->val);
inorder(vec,root->right);
}
};
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
stack<TreeNode*> tree_stack;
vector<int> vec;
TreeNode *cur_node = root;
while(cur_node||(!tree_stack.empty())){
while(cur_node){
tree_stack.push(cur_node);
cur_node = cur_node->left;
}
if(!tree_stack.empty()){
cur_node = tree_stack.top();
tree_stack.pop();
vec.push_back(cur_node->val);
cur_node = cur_node->right;
}
}
return vec;
}
};
来源:CSDN
作者:南方铁匠
链接:https://blog.csdn.net/hit_shaoqi/article/details/104427260