title
Given a binary tree, return the postorder traversal of its nodes’ values.
For example:
Given binary tree{1,#,2,3},
1
\
2
/
3
return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?
二叉树后序非递归遍历
solution
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayDeque<TreeNode> stack=new ArrayDeque<>();
if(root==null) return new ArrayList<>();
stack.push(root);
ArrayList<Integer> res=new ArrayList<>();
TreeNode pre=null,cur=null;
while (!stack.isEmpty()){
cur=stack.peek();
if(cur.left==null&&cur.right==null||(pre!=null&&(pre==cur.left||pre==cur.right))) {
res.add(cur.val);
stack.pop();
pre=cur;
}else{
if(cur.right!=null){
stack.push(cur.right);
}
if(cur.left!=null){
stack.push(cur.left);
}
}
}
return res;
}
}