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145. Binary Tree Postorder Traversal

Description

Given the root of a binary tree, return the postorder traversal of its nodes' values.

 

Example 1:

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

1. Recusive Traversal

2. Non-recursive using Stack

3. Morris Traversal

  • /**
 * 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;
    }
}

  • /**
 * 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;
    }
};

  • # 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]


  • /**
 * 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
}

  • /**
 * 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;
}


  • // 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
    }
}

All Problems

All Solutions