LeetCode 144. Binary Tree Preorder Traversal題目分析代碼

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?

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題目

給出一棵二叉樹，傳回其節點值的前序周遊。

分析

前序排序的非遞歸算法是相對簡單的，隻要使用棧，保證right，left的順序

代碼

方法一，非遞歸，運用一個棧即可

/**
 * 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
        Stack<TreeNode> stack = new Stack<TreeNode>();
        ArrayList<Integer> preorder = new ArrayList<Integer>();
        
        if (root == null) {
            return preorder;
        }
        
        stack.push(root);
        while (!stack.empty()) {
            TreeNode node = stack.pop();
            preorder.add(node.val);
            if (node.right != null) {
                stack.push(node.right);
            }
            if (node.left != null) {
                stack.push(node.left);
            }
        }
        
        return preorder;
    }
    
}           

複制

方法二： 遞歸周遊

public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Traversal(root,res);
        return res;
    }
    
    private void Traversal(TreeNode root, List<Integer> res) {
        if(root == null)
            return;
        res.add(root.val);
        Traversal(root.left,res);
        Traversal(root.right,res);
    }           

複制

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方法三 ： 分治法

public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null)
            return res;
        
        res.add(root.val);
        List<Integer> left = preorderTraversal(root.left);
        List<Integer> right = preorderTraversal(root.right);
        
        res.addAll(left);
        res.addAll(right);
        return res;
    }           

複制

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