LeetCode94 Binary Tree Inorder Traversal 二叉树中序遍历

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [1,3,2]

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

题源：here；完整实现：here

思路：

递归和迭代两种版本

1 递归

void inorder(TreeNode *t, vector<int> &result){
	if (!t) return;
	inorder(t->left, result);
	result.push_back(t->val);
	inorder(t->right, result);		
}
vector<int> inorderTraversal(TreeNode* root) {
	vector<int> result;
	inorder(root, result);
	return result;
}

纪念贴图：

2 迭代

vector<int> inorderTraversal2(TreeNode* root){
	stack<TreeNode*> records;
	vector<int> result;
	TreeNode *curr = root;
	while (curr || !records.empty()){
		while (curr){
			records.push(curr);
			curr = curr->left;
		}
		curr = records.top(); records.pop();
		result.push_back(curr->val);
		curr = curr->right;
	}

	return result;
}

纪念贴图：