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; } }
leetcode 1145. Binary Tree Postorder Traversal ----- java
原文：http://www.cnblogs.com/xiaoba1203/p/6068827.html