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144. Binary Tree Preorder Traversal

Description

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

 

Example 1:

Input: root = [1,null,2,3]
Output: [1,2,3]

Example 2:

Input: root = []
Output: []

Example 3:

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

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> preorderTraversal(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) {
                    ans.add(root.val);
                    prev.right = root;
                    root = root.left;
                } else {
                    prev.right = null;
                    root = root.right;
                }
            }
        }
        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> preorderTraversal(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) {
                    ans.push_back(root->val);
                    prev->right = root;
                    root = root->left;
                } else {
                    prev->right = nullptr;
                    root = root->right;
                }
            }
        }
        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 preorderTraversal(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:
                    ans.append(root.val)
                    prev.right = root
                    root = root.left
                else:
                    prev.right = None
                    root = root.right
        return ans


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) []int {
	var 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 {
				ans = append(ans, root.Val)
				prev.Right = root
				root = root.Left
			} else {
				prev.Right = nil
				root = root.Right
			}
		}
	}
	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 preorderTraversal(root: TreeNode | null): number[] {
    let 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) {
                ans.push(root.val);
                prev.right = root;
                root = root.left;
            } else {
                prev.right = null;
                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();
        res.push(node.val);
        Self::dfs(&node.left, res);
        Self::dfs(&node.right, res);
    }

    pub fn preorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = vec![];
        Self::dfs(&root, &mut res);
        res
    }
}

All Problems

All Solutions