题解 | #二叉树的后序遍历#
二叉树的后序遍历
http://www.nowcoder.com/practice/1291064f4d5d4bdeaefbf0dd47d78541
先处理根节点比较方便
import java.util.*;
/*
* public class TreeNode {
* int val = 0;
* TreeNode left = null;
* TreeNode right = null;
* public TreeNode(int val) {
* this.val = val;
* }
* }
*/
public class Solution {
/**
* 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
*
*
* @param root TreeNode类
* @return int整型一维数组
*/
public int[] postorderTraversal (TreeNode root) {
// write code here
if(root==null) return new int[0];
//两个栈实现:左右根——》根右左
Stack<TreeNode> sk1 = new Stack<>();
Stack<TreeNode> sk2 = new Stack<>();
sk1.push(root);
while(!sk1.isEmpty()){
root = sk1.pop();
sk2.push(root);
if(root.left!=null) sk1.push(root.left);
if(root.right!=null) sk1.push(root.right);
}
int len = sk2.size();
int[] result = new int[len];
for(int i = 0;i < len;i++){
result[i] = sk2.pop().val;
}
return result;
}
}