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问题陈述： 给定一棵二叉树，实现中序遍历并返回包含其中序序列的数组
例如给定下列二叉树：
我们按照左、根、右的顺序递归遍历二叉树，得到以下遍历：
最终中序遍历结果可以输出为： [3, 1, 9, 2, 4, 7, 5, 8, 6]
Morris trick
Morris 中序遍历是一种树遍历算法，旨在实现 O(1) 的空间复杂度，无需递归或外部数据结构。该算法应高效地按中序顺序访问二叉树中的每个节点，并在遍历过程中打印或处理节点值，而无需使用堆栈或递归。
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#include <iostream>
#include <sstream>
#include <unordered_map>
#include <vector>
#include <queue>
#include <map>
using namespace std;
// TreeNode structure
struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};
class Solution {
public:
// Function to perform iterative Morris
// inorder traversal of a binary tree
vector<int> getInorder(TreeNode* root) {
// Vector to store the
// inorder traversal result
vector<int> inorder;
// Pointer to the current node,
// starting from the root
TreeNode* cur = root;
// Loop until the current
// node is not NULL
while (cur != NULL) {
// If the current node's
// left child is NULL
if (cur->left == NULL) {
// Add the value of the current
// node to the inorder vector
inorder.push_back(cur->val);
// Move to the right child
cur = cur->right;
} else {
// If the left child is not NULL,
// find the predecessor (rightmost node
// in the left subtree)
TreeNode* prev = cur->left;
while (prev->right && prev->right != cur) {
prev = prev->right;
}
// If the predecessor's right child
// is NULL, establish a temporary link
// and move to the left child
if (prev->right == NULL) {
prev->right = cur;
cur = cur->left;
} else {
// If the predecessor's right child
// is already linked, remove the link,
// add current node to inorder vector,
// and move to the right child
prev->right = NULL;
inorder.push_back(cur->val);
cur = cur->right;
}
}
}
// Return the inorder
// traversal result
return inorder;
}
};
int main() {
TreeNode* root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->left = new TreeNode(4);
root->left->right = new TreeNode(5);
root->left->right->right = new TreeNode(6);
Solution sol;
vector<int> inorder = sol.getInorder(root);
cout << "Binary Tree Morris Inorder Traversal: ";
for(int i = 0; i< inorder.size(); i++){
cout << inorder[i] << " ";
}
cout << endl;
return 0;
}