Description
Given the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3]
Output: [1,3,2]
Explanation:
Example 2:
Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]
Output: [4,2,6,5,7,1,3,9,8]
Explanation:
Example 3:
Input: root = []
Output: []
Example 4:
Input: root = [1]
Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100].
-100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
Solutions
Solution 1: Recursive Traversal
We first recursively traverse the left subtree, then visit the root node, and finally recursively traverse the right subtree.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree, and the space complexity mainly depends on the stack space of the recursive call.
PythonJavaC++GoTypeScriptRustJavaScript
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def dfs(root):
if root is None:
return
dfs(root.left)
ans.append(root.val)
dfs(root.right)
ans = []
dfs(root)
return ans(code-box)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private List<Integer> ans = new ArrayList<>();
public List<Integer> inorderTraversal(TreeNode root) {
dfs(root);
return ans;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
ans.add(root.val);
dfs(root.right);
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ans;
function<void(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return;
}
dfs(root->left);
ans.push_back(root->val);
dfs(root->right);
};
dfs(root);
return ans;
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func inorderTraversal(root *TreeNode) (ans []int) {
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
ans = append(ans, root.Val)
dfs(root.Right)
}
dfs(root)
return
}(code-box)
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function inorderTraversal(root: TreeNode | null): number[] {
const ans: number[] = [];
const dfs = (root: TreeNode | null) => {
if (!root) {
return;
}
dfs(root.left);
ans.push(root.val);
dfs(root.right);
};
dfs(root);
return ans;
}(code-box)
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, ans: &mut Vec<i32>) {
if root.is_none() {
return;
}
let node = root.as_ref().unwrap().borrow();
Self::dfs(&node.left, ans);
ans.push(node.val);
Self::dfs(&node.right, ans);
}
pub fn inorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut ans = vec![];
Self::dfs(&root, &mut ans);
ans
}
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function (root) {
const ans = [];
const dfs = root => {
if (!root) {
return;
}
dfs(root.left);
ans.push(root.val);
dfs(root.right);
};
dfs(root);
return ans;
};(code-box)
Solution 2: Stack Implementation for Non-recursive Traversal
The non-recursive approach is as follows:
- Define a stack
stk.
- Push the left nodes of the tree into the stack in sequence.
- When the left node is null, pop and process the top element of the stack.
- Repeat steps 2-3.
The time complexity is O(n), and the space complexity is O(n). Here, n is the number of nodes in the binary tree, and the space complexity mainly depends on the stack space.
PythonJavaC++GoTypeScriptRustJavaScript
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans, stk = [], []
while root or stk:
if root:
stk.append(root)
root = root.left
else:
root = stk.pop()
ans.append(root.val)
root = root.right
return ans(code-box)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<>();
Deque<TreeNode> stk = new ArrayDeque<>();
while (root != null || !stk.isEmpty()) {
if (root != null) {
stk.push(root);
root = root.left;
} else {
root = stk.pop();
ans.add(root.val);
root = root.right;
}
}
return ans;
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ans;
stack<TreeNode*> stk;
while (root || stk.size()) {
if (root) {
stk.push(root);
root = root->left;
} else {
root = stk.top();
stk.pop();
ans.push_back(root->val);
root = root->right;
}
}
return ans;
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func inorderTraversal(root *TreeNode) (ans []int) {
stk := []*TreeNode{}
for root != nil || len(stk) > 0 {
if root != nil {
stk = append(stk, root)
root = root.Left
} else {
root = stk[len(stk)-1]
stk = stk[:len(stk)-1]
ans = append(ans, root.Val)
root = root.Right
}
}
return
}(code-box)
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function inorderTraversal(root: TreeNode | null): number[] {
const stk: TreeNode[] = [];
const ans: number[] = [];
while (root || stk.length > 0) {
if (root) {
stk.push(root);
root = root.left;
} else {
root = stk.pop();
ans.push(root.val);
root = root.right;
}
}
return ans;
}(code-box)
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn inorder_traversal(mut root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut ans = vec![];
let mut stk = vec![];
while root.is_some() || !stk.is_empty() {
if root.is_some() {
let next = root.as_mut().unwrap().borrow_mut().left.take();
stk.push(root);
root = next;
} else {
let mut node = stk.pop().unwrap();
let mut node = node.as_mut().unwrap().borrow_mut();
ans.push(node.val);
root = node.right.take();
}
}
ans
}
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function (root) {
const stk = [];
const ans = [];
while (root || stk.length > 0) {
if (root) {
stk.push(root);
root = root.left;
} else {
root = stk.pop();
ans.push(root.val);
root = root.right;
}
}
return ans;
};(code-box)
Solution 3: Morris Implementation for In-order Traversal
Morris traversal does not require a stack, so the space complexity is O(1). The core idea is:
Traverse the binary tree nodes,
- If the left subtree of the current node
root is null, add the current node value to the result list ans, and update the current node to root.right.
- If the left subtree of the current node
root is not null, find the rightmost node prev of the left subtree (which is the predecessor node of the root node in in-order traversal):
- If the right subtree of the predecessor node
prev is null, point the right subtree of the predecessor node to the current node root, and update the current node to root.left.
- If the right subtree of the predecessor node
prev is not null, add the current node value to the result list ans, then point the right subtree of the predecessor node to null (i.e., disconnect prev and root), and update the current node to root.right.
- Repeat the above steps until the binary tree node is null, and the traversal ends.
The time complexity is O(n), and the space complexity is O(1). Here, n is the number of nodes in the binary tree.
PythonJavaC++GoTypeScriptJavaScript
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
while root:
if root.left is None:
ans.append(root.val)
root = root.right
else:
prev = root.left
while prev.right and prev.right != root:
prev = prev.right
if prev.right is None:
prev.right = root
root = root.left
else:
ans.append(root.val)
prev.right = None
root = root.right
return ans(code-box)
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<>();
while (root != null) {
if (root.left == null) {
ans.add(root.val);
root = root.right;
} else {
TreeNode prev = root.left;
while (prev.right != null && prev.right != root) {
prev = prev.right;
}
if (prev.right == null) {
prev.right = root;
root = root.left;
} else {
ans.add(root.val);
prev.right = null;
root = root.right;
}
}
}
return ans;
}
}(code-box)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ans;
while (root) {
if (!root->left) {
ans.push_back(root->val);
root = root->right;
} else {
TreeNode* prev = root->left;
while (prev->right && prev->right != root) {
prev = prev->right;
}
if (!prev->right) {
prev->right = root;
root = root->left;
} else {
ans.push_back(root->val);
prev->right = nullptr;
root = root->right;
}
}
}
return ans;
}
};(code-box)
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func inorderTraversal(root *TreeNode) (ans []int) {
for root != nil {
if root.Left == nil {
ans = append(ans, root.Val)
root = root.Right
} else {
prev := root.Left
for prev.Right != nil && prev.Right != root {
prev = prev.Right
}
if prev.Right == nil {
prev.Right = root
root = root.Left
} else {
ans = append(ans, root.Val)
prev.Right = nil
root = root.Right
}
}
}
return
}(code-box)
/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function inorderTraversal(root: TreeNode | null): number[] {
const ans: number[] = [];
while (root) {
if (!root.left) {
ans.push(root.val);
root = root.right;
} else {
let prev = root.left;
while (prev.right && prev.right != root) {
prev = prev.right;
}
if (!prev.right) {
prev.right = root;
root = root.left;
} else {
ans.push(root.val);
prev.right = null;
root = root.right;
}
}
}
return ans;
}(code-box)
/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
var inorderTraversal = function (root) {
const ans = [];
while (root) {
if (!root.left) {
ans.push(root.val);
root = root.right;
} else {
let prev = root.left;
while (prev.right && prev.right != root) {
prev = prev.right;
}
if (!prev.right) {
prev.right = root;
root = root.left;
} else {
ans.push(root.val);
prev.right = null;
root = root.right;
}
}
}
return ans;
};(code-box)
