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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].
牛解法：postorder就是L-R-ROOT，这里先ROOT-R-L，再把结果reverse
public class Solution { public List<Integer> postorderTraversal(TreeNode root) { List<Integer> results = new ArrayList<Integer>(); Deque<TreeNode> stack = new ArrayDeque<TreeNode>(); while (!stack.isEmpty() || root != null) { if (root != null) { stack.push(root); results.add(root.val); root = root.right; } else { root = stack.pop().left; } } Collections.reverse(results); return results; } }
关于deque：https://docs.oracle.com/javase/7/docs/api/java/util/Deque.html
reference：https://leetcode.com/discuss/9736/accepted-code-with-explaination-does-anyone-have-better-idea
**Binary Tree Postorder Traversal
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原文地址：http://www.cnblogs.com/hygeia/p/5101915.html
