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Description
给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/binary-tree-postorder-traversal
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
代码1 (java) 2020-09-29 -_-
/**
* 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 { //后序遍历 左树————右树————根节点
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>(); //初始化 返回值
postorder(root,res); //调用递归函数
return res;
}
public void postorder(TreeNode root, List<Integer> res){
if(root==null){ //判断没有节点时返回上一层
return;
}
postorder(root.left,res); //先遍历左树
postorder(root.right,res); //后遍历右树
res.add(root.val); //记录当前的值
}
}
//执行用时: 0 ms
//内存消耗: 37.4 MB
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