给你二叉树的根节点 root
,返回它节点值的 前序 遍历。
示例 1:
输入:root = [1,null,2,3]
输出:[1,2,3]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
示例 4:
输入:root = [1,2]
输出:[1,2]
示例 5:
输入:root = [1,null,2]
输出:[1,2]
提示:
树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100
进阶:递归算法很简单,你可以通过迭代算法完成吗?
1 /** 2 * Definition for a binary tree node. 3 * function TreeNode(val, left, right) { 4 * this.val = (val===undefined ? 0 : val) 5 * this.left = (left===undefined ? null : left) 6 * this.right = (right===undefined ? null : right) 7 * } 8 */ 9 /** 10 * @param {TreeNode} root 11 * @return {number[]} 12 */ 13 var preorderTraversal = function(root) { 14 const res = []; 15 const stack = []; //新建一个栈 16 if(root) stack.push(root); //根结点存在则入栈 17 while (stack.length) { 18 const n = stack.pop(); 19 res.push(n.val); 20 if(n.right) stack.push(n.right); //右先入栈,左后入,出栈的时候正好反过来,符合前序遍历的顺序 21 if(n.left) stack.push(n.left); 22 } 23 return res; 24 };