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144. Binary Tree Preorder Traversal

DifficultyEasy

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [1,2,4,5,6,7,3,8,9]

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

Thinking

Preorder is root, left, right. A tree is recursive, so record the root and recurse on both children. \(n\le 100\), so the call stack is fine. The follow-up asks for iteration.

We first visit the root node, then recursively traverse the left and right subtrees.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree. The space complexity mainly depends on the stack space used for recursive calls.

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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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root):
            if root is None:
                return
            ans.append(root.val)
            dfs(root.left)
            dfs(root.right)

        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 = new ArrayList<>();

    public List<Integer> preorderTraversal(TreeNode root) {
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        ans.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }
}

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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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            ans.push_back(root->val);
            dfs(root->left);
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        ans = append(ans, root.Val)
        dfs(root.Left)
        dfs(root.Right)
    }
    dfs(root)
    return
}

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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 preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        ans.push(root.val);
        dfs(root.left);
        dfs(root.right);
    };
    dfs(root);
    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::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();
        ans.push(node.val);
        Self::dfs(&node.left, ans);
        Self::dfs(&node.right, ans);
    }

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

Solution 2: Stack Implementation for Non-Recursive Traversal

Thinking

Solution 1 is already correct; the follow-up drops the call stack. An explicit stack pushes right then left, so popping visits root first. Same order, still \(O(n)\) space.

The idea of using a stack to implement non-recursive traversal is as follows:

  1. Define a stack \(stk\), and first push the root node into the stack.
  2. If the stack is not empty, pop a node from the stack each time.
  3. Process the node.
  4. First push the right child of the node into the stack, then push the left child of the node into the stack (if there are child nodes).
  5. Repeat steps 2-4.
  6. Return the result.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree. The space complexity mainly depends on the stack space.

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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 preorderTraversal(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.right:
                stk.append(node.right)
            if node.left:
                stk.append(node.left)
        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 {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> stk = new ArrayDeque<>();
        stk.push(root);
        while (!stk.isEmpty()) {
            TreeNode node = stk.pop();
            ans.add(node.val);
            if (node.right != null) {
                stk.push(node.right);
            }
            if (node.left != null) {
                stk.push(node.left);
            }
        }
        return ans;
    }
}

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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> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        if (!root) {
            return ans;
        }
        stack<TreeNode*> stk;
        stk.push(root);
        while (stk.size()) {
            auto node = stk.top();
            stk.pop();
            ans.push_back(node->val);
            if (node->right) {
                stk.push(node->right);
            }
            if (node->left) {
                stk.push(node->left);
            }
        }
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(root *TreeNode) (ans []int) {
    if root == nil {
        return
    }
    stk := []*TreeNode{root}
    for len(stk) > 0 {
        node := stk[len(stk)-1]
        stk = stk[:len(stk)-1]
        ans = append(ans, node.Val)
        if node.Right != nil {
            stk = append(stk, node.Right)
        }
        if node.Left != nil {
            stk = append(stk, node.Left)
        }
    }
    return
}

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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 preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    if (!root) {
        return ans;
    }
    const stk: TreeNode[] = [root];
    while (stk.length) {
        const { left, right, val } = stk.pop();
        ans.push(val);
        right && stk.push(right);
        left && stk.push(left);
    }
    return ans;
}

Solution 3: Morris Preorder Traversal

Thinking

Solution 2 still uses \(O(h)\) space. Morris temporarily links the rightmost node of the left subtree to the current root, emits the root on the first visit, and clears the link on return. The tree's null pointers act as the stack, \(O(1)\) space.

Morris traversal does not require a stack, and its space complexity is \(O(1)\). The core idea is:

Traverse the binary tree nodes,

  1. If the left subtree of the current node root is empty, add the current node value to the result list \(ans\), and update the current node to root.right.
  2. If the left subtree of the current node root is not empty, find the rightmost node pre of the left subtree (which is the predecessor of the root node in inorder traversal):
    • If the right subtree of the predecessor node pre is empty, add the current node value to the result list \(ans\), then 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 pre is not empty, point the right subtree of the predecessor node to null (i.e., disconnect pre and root), and update the current node to root.right.
  3. Repeat the above steps until the binary tree node is null, and the traversal ends.

The time complexity is \(O(n)\), where \(n\) is the number of nodes in the binary tree. The space complexity is \(O(1)\).

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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 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

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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> 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;
    }
}

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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> 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;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func preorderTraversal(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 {
                ans = append(ans, root.Val)
                prev.Right = root
                root = root.Left
            } else {
                prev.Right = nil
                root = root.Right
            }
        }
    }
    return
}

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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 preorderTraversal(root: TreeNode | null): number[] {
    const ans: number[] = [];
    while (root) {
        const { left, right, val } = root;
        if (!left) {
            ans.push(val);
            root = right;
        } else {
            let prev = left;
            while (prev.right && prev.right != root) {
                prev = prev.right;
            }
            if (!prev.right) {
                ans.push(val);
                prev.right = root;
                root = root.left;
            } else {
                prev.right = null;
                root = root.right;
            }
        }
    }
    return ans;
}

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