【LeetCode】94. Binary Tree Inorder Traversal（C++）

地址：https://leetcode.com/problems/binary-tree-inorder-traversal/

题目：

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

Example:

理解：

二叉树的中序遍历

递归实现：

class Solution {
public:
	vector<int> inorderTraversal(TreeNode* root) {
		vector<int> res;
		InOrder(root, res);
		return res;
	}

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

非递归实现：

class Solution {
public:
	vector<int> inorderTraversal(TreeNode* root) {
		vector<int> res;
		stack<TreeNode*> stk;
		TreeNode* p = root;
		while (p || !stk.empty()) {
			if (p) {
				stk.push(p);
				p = p->left;
			}
			else {
				p= stk.top();
				stk.pop();
				res.push_back(p->val);
				p = p->right;
			}
		}
		return res;
	}
};

基于线索树的实现

前面两种空间复杂度都是 $O(n)$
基于线索数，可以把空间复杂度变为 $O(1)$

class Solution {
public:
	vector<int> inorderTraversal(TreeNode* root) {
		vector<int> res;
		TreeNode* curr = root;
		while (curr) {
			if (curr->left) {
				TreeNode* pre = curr->left;
				while (pre->right && (pre->right != curr)) {
					pre = pre->right;
				}
				//如果是第一次到达，修改线索树
				if (!pre->right) {
					pre->right = curr;
					curr = curr->left;
				}
				//第二次到达，改回去
				else {
					pre->right = nullptr;
					res.push_back(curr->val);
					curr = curr->right;
				}
			}
			else {
				res.push_back(curr->val);
				curr = curr->right;
			}
		}
		return res;
	}
};