144.Binary Tree Preorder Traversal

tags: Easy,Tree,DFS,Stack,Binary Tree

144. Binary Tree Preorder Traversal

題目描述

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

範例

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?

解答

Javascript

recursive

function preorderTraversal(root) {
  const result = [];
  if (root === null) return result;
  result.push(root.val);
  result.push(...preorderTraversal(root.left));
  result.push(...preorderTraversal(root.right));
  return result;
}

iterative

function preorderTraversal2(root) {
  const result = [];
  if (root === null) return result;
  const stack = [root];
  while (stack.length > 0) {
    const node = stack.pop();
    result.push(node.val);
    if (node.right !== null) stack.push(node.right);
    if (node.left !== null) stack.push(node.left);
  }

  return result;
}

Marsgoat Jan 9, 2023

Python

class Solution:
    def preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
        if not root: return root
        
        traversal, stk = [], [root]

        while stk:
            node = stk.pop()
            traversal.append(node.val)

            if node.right:
                stk.append(node.right)
            if node.left:
                stk.append(node.left)

        return traversal

Ron Chen Jan 9, 2023

Reference

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