[leecode]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?

算法思路：

思路1：递归版

代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> res = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {
        if(root == null) return res;
        res.add(root.val);
        if(root.left != null) preorderTraversal(root.left);
        if(root.right != null) preorderTraversal(root.right);
        return res;
    }
}

2. 非递归实现：

借助栈，每一次弹栈处理栈顶元素之后都将右节点、左节点分别压栈，大家画个图就明白了

代码如下：

public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            res.add(node.val);
            if(node.right != null){
                stack.push(node.right);
            }
            if(node.left != null){
                stack.push(node.left);
            }
        }
        return res;
    }
}