leetcode 094 Binary Tree Inorder Traversal

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本文介绍了一种实现二叉树中序遍历的方法，包括递归与迭代两种方式。递归方法直观易懂，而迭代方法使用栈来辅助完成，避免了递归带来的栈溢出风险。

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

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

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class Solution {
public:
	void inOrder(TreeNode* root, vector<int> &res) {
		if(root!=NULL) {
			inOrder(root->left, res);
			res.push_back(root->val);
			inOrder(root->right, res);
		}
		return;
	}
	vector<int> inorderTraversal(TreeNode* root) {
		vector<int> res;
		stack<TreeNode*> s;
		if(root==NULL) return res;

		s.push(root);

		while(!s.empty()) {
			TreeNode *temp = s.top();

			if(temp->left!=NULL) {
				s.push(temp->left);
				temp->left=NULL;
			} else {
				s.pop();
				res.push_back(temp->val);
				if(temp->right!=NULL) {
					s.push(temp->right);
				}
			}
		}
		return res;
	}
}