题目大意
https://leetcode.com/problems/binary-tree-inorder-traversal/description/
94. Binary Tree Inorder Traversal
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?
解题思路
中序遍历：左根右
Approach 1: Recursive Approach 递归
The first method to solve this problem is using recursion. This is the classical method and is straightforward. We can define a helper function to implement recursion.
Python解法
class Solution(object): def inorderTraversal(self, root): # 递归 """ :type root: TreeNode :rtype: List[int] """ if not root: return [] return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right)
Java解法
class Solution { public List < Integer > inorderTraversal(TreeNode root) { List < Integer > res = new ArrayList < > (); helper(root, res); return res; } public void helper(TreeNode root, List < Integer > res) { if (root != null) { if (root.left != null) { helper(root.left, res); } res.add(root.val); if (root.right != null) { helper(root.right, res); } } } }
Complexity Analysis
Time complexity : O(n)O(n). The time complexity is O(n)O(n) because the recursive function is T(n) = 2 cdot T(n/2)+1T(n)=2⋅T(n/2)+1.
Space complexity : The worst case space required is O(n)O(n), and in the average case it's O(log(n))O(log(n)) where nn is number of nodes.
Approach 2: Iterating method using Stack 迭代（基于栈）
The strategy is very similiar to the first method, the different is using stack.
伪代码如下（摘录自Wikipedia Tree_traversal）
iterativeInorder(node) parentStack = empty stack while (not parentStack.isEmpty() or node ≠ null) if (node ≠ null) parentStack.push(node) node = node.left else node = parentStack.pop() visit(node) node = node.right
Python解法
class Solution(object): def inorderTraversal(self, root): # 迭代 """ :type root: TreeNode :rtype: List[int] """ stack = [] res = [] while root or stack: while root: stack.append(root) root = root.left root = stack.pop() res.append(root.val) root = root.right return res
Java解法
public class Solution { public List < Integer > inorderTraversal(TreeNode root) { List < Integer > res = new ArrayList < > (); Stack < TreeNode > stack = new Stack < > (); TreeNode curr = root; while (curr != null || !stack.isEmpty()) { while (curr != null) { stack.push(curr); curr = curr.left; } curr = stack.pop(); res.add(curr.val); curr = curr.right; } return res; } }
Complexity Analysis
Time complexity : O(n)O(n).
Space complexity : O(n)O(n).
Approach 3: Morris Traversal
In this method, we have to use a new data structure-Threaded Binary Tree, and the strategy is as follows:
Step 1: Initialize current as root
Step 2: While current is not NULL,
If current does not have left child
a. Add current’s value
b. Go to the right, i.e., current = current.right
Else
a. In current's left subtree, make current the right child of the rightmost node
b. Go to this left child, i.e., current = current.left
For example:
1
/
2 3
/ /
4 5 6
First, 1 is the root, so initialize 1 as current, 1 has left child which is 2, the current's left subtree is
2
/
4 5
So in this subtree, the rightmost node is 5, then make the current(1) as the right child of 5. Set current = cuurent.left (current = 2). The tree now looks like:
2
/
4 5
1
3
/
6
For current 2, which has left child 4, we can continue with thesame process as we did above
4
2
5
1
3
/
6
then add 4 because it has no left child, then add 2, 5, 1, 3 one by one, for node 3 which has left child 6, do the same as above. Finally, the inorder taversal is [4,2,5,1,6,3].
For more details, please check Threaded binary tree and Explaination of Morris Method
Python解法
class Solution(object): def inorderTraversal(self, root): # Morris Traversal """ :type root: TreeNode :rtype: List[int] """ res = [] curr, pre = root, None while curr: if curr.left: pre = curr.left while pre.right: pre = pre.right pre.right = curr curr.left, curr = None, curr.left else: res.append(curr.val) curr = curr.right return res
Java解法
class Solution { public List < Integer > inorderTraversal(TreeNode root) { List < Integer > res = new ArrayList < > (); TreeNode curr = root; TreeNode pre; while (curr != null) { if (curr.left == null) { res.add(curr.val); curr = curr.right; // move to next right node } else { // has a left subtree pre = curr.left; while (pre.right != null) { // find rightmost pre = pre.right; } pre.right = curr; // put cur after the pre node TreeNode temp = curr; // store cur node curr = curr.left; // move cur to the top of the new tree temp.left = null; // original cur left be null, avoid infinite loops } } return res; } }
Complexity Analysis
Time complexity : O(n)O(n). To prove that the time complexity is O(n)O(n), the biggest problem lies in finding the time complexity of finding the predecessor nodes of all the nodes in the binary tree. Intuitively, the complexity is O(nlogn)O(nlogn), because to find the predecessor node for a single node related to the height of the tree. But in fact, finding the predecessor nodes for all nodes only needs O(n)O(n) time. Because a binary Tree with nn nodes has n-1n−1 edges, the whole processing for each edges up to 2 times, one is to locate a node, and the other is to find the predecessor node. So the complexity is O(n)O(n).
Space complexity : O(n)O(n). Arraylist of size nn is used.
参考：
https://leetcode.com/problems/binary-tree-inorder-traversal/solution/
http://bookshadow.com/weblog/2015/01/19/leetcode-binary-tree-inorder-traversal/