Binary Tree Preorder Traversal

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

Given:

    1
   /   2   3
 / 4   5

return [1,2,4,5,3].

Thinking:

For this problem, you need to think about using recursive or non-recursive methods. As recursive method, we should think the order of traverse is to visit the node itself then visit left and right subtrees for traverse. So use a helper method to pass the list into which to acomplish hte adding of numbers to hte list.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root == null) {
            return result;
        }
        traverse(root,result);
        return result;
    }
    public void traverse (TreeNode root,ArrayList<Integer> list) {
        if (root == null) {
            return;
        }
        list.add(root.val);
        traverse(root.left,list);
        traverse(root.right,list);
    }
}

For non-recursive method, think about using a stack to push nodes in order of right first then left. Because this order will lead to right node is always below the left node. Meanwhile, before finishing the left node‘s children and decedents you will go deep until leaves.

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Preorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> preorderTraversal(TreeNode root) {
        // write your code here
        // use stack to achieve pre-order
        Stack<TreeNode> stack = new Stack<TreeNode>();
        ArrayList<Integer> result = new ArrayList<Integer>();
        stack.push(root);
        // eadge case
        if (root == null) {
            return result;
        }
        // while loop to traverse every node in tree
        while (!stack.isEmpty()) {
            TreeNode curr = stack.pop();
            result.add(curr.val);
            if (curr.right != null) {
                stack.push(curr.right);
            }
            if (curr.left != null) {
                stack.push(curr.left);
            }
        }

        return result;
    }

}
时间: 2024-10-09 21:50:51

Binary Tree Preorder Traversal的相关文章

[LeetCode][JavaScript]Binary Tree Preorder Traversal

Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? https://leetcod

LeetCode: Binary Tree Preorder Traversal 解题报告

Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3},   1    \     2    /   3return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively?

40: Binary Tree Preorder Traversal

/************************************************************************/            /*       40:  Binary Tree Preorder Traversal                               */            /************************************************************************/ 

LeetCode:Binary Tree Preorder Traversal

题目:Binary Tree Preorder Traversal 二叉树的前序遍历,同样使用栈来解,代码如下: 1 struct TreeNode { 2 int val; 3 TreeNode* left; 4 TreeNode* right; 5 TreeNode(int x): val(x), left(NULL),right(NULL) {} 6 }; 7 8 vector<int> preorderTraversal(TreeNode *root) //非递归的前序遍历(用栈实现)

[LeetCode 题解]: Binary Tree Preorder Traversal

Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 题意 先序遍历二叉树,递归的思路是普通的,能否用迭代呢? 非递归思路:<借助stack>

LeetCode 144. Binary Tree Preorder Traversal 解题报告

144. Binary Tree Preorder Traversal My Submissions Question Total Accepted: 108336 Total Submissions: 278322 Difficulty: Medium Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3

Binary Tree Preorder Traversal and Binary Tree Postorder Traversal

Binary Tree Preorder Traversal Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 2 / 3 return [1,2,3]. c++版: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode

[Leetcode][JAVA] Binary Tree Preorder Traversal, Binary Tree Inorder Traversal, Binary Tree Postorder Traversal

Binary Tree PreOrder Traversal: Given a binary tree, return the preorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 2 / 3   return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 不使用递归前序遍历,可以

Binary Tree Postorder Traversal &amp;&amp; Binary Tree Preorder Traversal

详见:剑指 Offer 题目汇总索引:第6题 Binary Tree Postorder Traversal            Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 2 / 3 return [3,2,1]. Note: Recursive solution is trivial, could y

leetcode - Binary Tree Preorder Traversal &amp;&amp; Binary Tree Inorder Traversal &amp;&amp; Binary Tree Postorder Traversal

简单来说,就是二叉树的前序.中序.后序遍历,包括了递归和非递归的方法 前序遍历(注释中的为递归版本): 1 #include <vector> 2 #include <stack> 3 #include <stddef.h> 4 #include <iostream> 5 6 using namespace std; 7 8 struct TreeNode 9 { 10 int val; 11 TreeNode *left; 12 TreeNode *rig