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题目:(Tree ,Stack)
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]
.
题解:
binary tree 三部曲里面最难的一个。多看看。
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public ArrayList<Integer> postorderTraversal(TreeNode root) { ArrayList<Integer> result = new ArrayList<Integer>(); if(root==null) return result; Stack<TreeNode> stack=new Stack<TreeNode>(); stack.push(root); TreeNode prev=null; while(!stack.isEmpty()){ TreeNode current=stack.peek(); if(prev==null||prev.left==current||prev.right==current) { if(current.left!=null) stack.push(current.left); else if(current.right!=null) stack.push(current.right); else { stack.pop(); result.add(current.val); } }else if(prev==current.left) { if(current.right!=null) stack.push(current.right); else { stack.pop(); result.add(current.val); } }else { stack.pop(); result.add(current.val); } prev=current; } return result; } }
[leetcode] Binary Tree Postorder Traversal
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原文地址:http://www.cnblogs.com/fengmangZoo/p/4196980.html