2014
11-18

# LeetCode-Binary Tree Inorder Traversal[二叉树]

### Binary Tree Inorder Traversal

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

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

   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:

   1
/ \
2   3
/
4
\
5


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

// LeetCode, Binary Tree Inorder Traversal
// 使用栈，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> result;
const TreeNode *p = root;
stack<const TreeNode *> s;

while (!s.empty() || p != nullptr) {
if (p != nullptr) {
s.push(p);
p = p->left;
} else {
p = s.top();
s.pop();
result.push_back(p->val);
p = p->right;
}
}
return result;
}
};


// LeetCode, Binary Tree Inorder Traversal
// Morris中序遍历，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
vector<int> result;
TreeNode *cur, *prev;

cur = root;
while (cur != nullptr) {
if (cur->left == nullptr) {
result.push_back(cur->val);
prev = cur;
cur = cur->right;
} else {
/* 查找前驱 */
TreeNode *node = cur->left;
while (node->right != nullptr && node->right != cur)
node = node->right;

if (node->right == nullptr) { /* 还没线索化，则建立线索 */
node->right = cur;
/* prev = cur; 不能有这句，cur还没有被访问 */
cur = cur->left;
} else {    /* 已经线索化，则访问节点，并删除线索  */
result.push_back(cur->val);
node->right = nullptr;
prev = cur;
cur = cur->right;
}
}
}
return result;
}
};


Java代码:

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
while(root != null || stack.size() > 0){
while(root != null){

stack.push(root);
root = root.left;
}
if(stack.size() > 0){
TreeNode top = stack.pop();
}