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?
求后序遍历，要求不使用递归。
使用栈，从后向前添加。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List list = new ArrayList<Integer>(); if( root == null )
return list;
Stack<TreeNode> stack = new Stack<TreeNode>(); stack.push(root); while( !stack.isEmpty() ){ TreeNode node = stack.pop();
list.add(0,node.val);
if( node.left != null )
stack.push(node.left);
if( node.right != null )
stack.push(node.right); }
return list; }
}