Description#
Given the root
of a binary tree, return the postorder traversal of its nodes' values.
Example 1:
![](https://spcdn.pages.dev/leetcode/problems/0145.Binary%20Tree%20Postorder%20Traversal/images/pre1.jpg)
Input: root = [1,null,2,3]
Output: [3,2,1]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Constraints:
- The number of the 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#
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| # 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def dfs(root):
if root is None:
return
dfs(root.left)
dfs(root.right)
nonlocal ans
ans.append(root.val)
ans = []
dfs(root)
return ans
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| /**
* 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;
public List<Integer> postorderTraversal(TreeNode root) {
ans = new ArrayList<>();
dfs(root);
return ans;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
dfs(root.right);
ans.add(root.val);
}
}
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| /**
* 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> postorderTraversal(TreeNode* root) {
vector<int> ans;
while (root) {
if (!root->right) {
ans.push_back(root->val);
root = root->left;
} else {
TreeNode* next = root->right;
while (next->left && next->left != root) {
next = next->left;
}
if (!next->left) {
ans.push_back(root->val);
next->left = root;
root = root->right;
} else {
next->left = nullptr;
root = root->left;
}
}
}
reverse(ans.begin(), ans.end());
return ans;
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func postorderTraversal(root *TreeNode) []int {
var ans []int
for root != nil {
if root.Right == nil {
ans = append([]int{root.Val}, ans...)
root = root.Left
} else {
next := root.Right
for next.Left != nil && next.Left != root {
next = next.Left
}
if next.Left == nil {
ans = append([]int{root.Val}, ans...)
next.Left = root
root = root.Right
} else {
next.Left = nil
root = root.Left
}
}
}
return ans
}
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| /**
* 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 postorderTraversal(root: TreeNode | null): number[] {
if (root == null) return [];
let stack = [];
let ans = [];
let prev = null;
while (root || stack.length) {
while (root) {
stack.push(root);
root = root.left;
}
root = stack.pop();
if (!root.right || root.right == prev) {
ans.push(root.val);
prev = root;
root = null;
} else {
stack.push(root);
root = root.right;
}
}
return ans;
}
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| // 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut Vec<i32>) {
if root.is_none() {
return;
}
let node = root.as_ref().unwrap().borrow();
Self::dfs(&node.left, res);
Self::dfs(&node.right, res);
res.push(node.val);
}
pub fn postorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
Self::dfs(&root, &mut res);
res
}
}
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Solution 2#
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| # 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
if root is None:
return ans
stk = [root]
while stk:
node = stk.pop()
ans.append(node.val)
if node.left:
stk.append(node.left)
if node.right:
stk.append(node.right)
return ans[::-1]
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| /**
* 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> postorderTraversal(TreeNode root) {
LinkedList<Integer> ans = new LinkedList<>();
if (root == null) {
return ans;
}
Deque<TreeNode> stk = new ArrayDeque<>();
stk.push(root);
while (!stk.isEmpty()) {
TreeNode node = stk.pop();
ans.addFirst(node.val);
if (node.left != null) {
stk.push(node.left);
}
if (node.right != null) {
stk.push(node.right);
}
}
return ans;
}
}
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| /**
* 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 postorderTraversal(root: TreeNode | null): number[] {
if (root == null) {
return [];
}
const { val, left, right } = root;
return [...postorderTraversal(left), ...postorderTraversal(right), val];
}
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| // 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
pub fn postorder_traversal(mut root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
let mut res = vec![];
let mut stack = vec![];
while root.is_some() || !stack.is_empty() {
if root.is_some() {
let next = root.as_mut().unwrap().borrow_mut().left.take();
stack.push(root);
root = next;
} else {
root = stack.pop().unwrap();
let next = root.as_mut().unwrap().borrow_mut().right.take();
if next.is_some() {
stack.push(root);
} else {
res.push(root.as_ref().unwrap().borrow().val);
}
root = next;
}
}
res
}
}
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Solution 3#
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| # 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
ans = []
while root:
if root.right is None:
ans.append(root.val)
root = root.left
else:
next = root.right
while next.left and next.left != root:
next = next.left
if next.left != root:
ans.append(root.val)
next.left = root
root = root.right
else:
next.left = None
root = root.left
return ans[::-1]
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| /**
* 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> postorderTraversal(TreeNode root) {
LinkedList<Integer> ans = new LinkedList<>();
while (root != null) {
if (root.right == null) {
ans.addFirst(root.val);
root = root.left;
} else {
TreeNode next = root.right;
while (next.left != null && next.left != root) {
next = next.left;
}
if (next.left == null) {
ans.addFirst(root.val);
next.left = root;
root = root.right;
} else {
next.left = null;
root = root.left;
}
}
}
return ans;
}
}
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| /**
* 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 postorderTraversal(root: TreeNode | null): number[] {
const res = [];
while (root != null) {
const { val, left, right } = root;
if (right == null) {
res.push(val);
root = left;
} else {
let next = right;
while (next.left != null && next.left != root) {
next = next.left;
}
if (next.left == null) {
res.push(val);
next.left = root;
root = right;
} else {
next.left = null;
root = left;
}
}
}
return res.reverse();
}
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