Daily Leetcode 144. Binary Tree Preorder Traversal

2023 Dec LC challenge

相似題目

Binary Tree Inorder Traversal
Binary Tree Postorder Traversal
Binary Tree Preorder Traversal

preorder、postorder、inorder traversal 特性

前序 Preorder : root -> left -> right
後序 Postorder : left -> right -> root
中序 InOrder : right -> left -> right

My Code

94. Binary Tree Inorder Traversal

中序 InOrder : right -> left -> right

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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> inorderTraversal(TreeNode root) {
        //root.left - > root -> root.right
        List<Integer> result = new ArrayList<>();
        inorderTraversal(root, result);
        return result;
    }

        private void inorderTraversal(TreeNode node, List<Integer> result) {
        if (node == null) return;

        if (node.left != null) {
            inorderTraversal(node.left, result);
        }

        result.add(node.val);

        if (node.right != null) {
            inorderTraversal(node.right, result);
        }
    }
}

144. Binary Tree Preorder Traversal

前序 Preorder : root -> left -> right

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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> result = new ArrayList<>();
        preorderTraversal(result, root);
        return result;
    }

    public void preorderTraversal(List<Integer> result, TreeNode node) {
        
        if ( node == null) return;
        
        result.add(node.val);

        if (node.left != null) {
            preorderTraversal(result, node.left);
        }

        if(node.right != null) {
            preorderTraversal(result, node.right);
        }
    } 
}

145. Binary Tree Postorder Traversal

後序 Postorder : left -> right -> root

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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) {
        
        List<Integer> result = new ArrayList<>();
        postorderTraversal(result, root);
        return result;
    }

    public void postorderTraversal(List<Integer> result, TreeNode node) {
        if (node == null) return;

        if (node.left != null) {
            postorderTraversal(result, node.left);
        }

        if (node.right != null) {
            postorderTraversal(result, node.right);
        }
        result.add(node.val);
    }
}