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Binary Tree Postorder Traversal
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?
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SOLUTION 1:
递归解法
1 public List<Integer> postorderTraversal1(TreeNode root) { 2 List<Integer> ret = new ArrayList<Integer>(); 3 dfs(root, ret); 4 return ret; 5 } 6 7 // Solution 1: rec 8 public void dfs(TreeNode root, List<Integer> ret) { 9 if (root == null) { 10 return; 11 } 12 13 dfs(root.left, ret); 14 dfs(root.right, ret); 15 ret.add(root.val); 16 }
SOLUTION 2:
/**
* 后序遍历迭代解法
* http://www.youtube.com/watch?v=hv-mJUs5mvU
* http://blog.csdn.net/tang_jin2015/article/details/8545457
* 从左到右的后序 与从右到左的前序的逆序是一样的，所以就简单喽！ 哈哈
* 用另外一个栈进行翻转即可喽
*/
1 // Solution 2: iterator 2 public List<Integer> postorderTraversal(TreeNode root) { 3 List<Integer> ret = new ArrayList<Integer>(); 4 if (root == null) { 5 return ret; 6 } 7 8 Stack<TreeNode> s = new Stack<TreeNode>(); 9 Stack<Integer> out = new Stack<Integer>(); 10 11 s.push(root); 12 13 while (!s.isEmpty()) { 14 TreeNode cur = s.pop(); 15 out.push(cur.val); 16 17 if (cur.left != null) { 18 s.push(cur.left); 19 } 20 21 if (cur.right != null) { 22 s.push(cur.right); 23 } 24 } 25 26 while (!out.isEmpty()) { 27 ret.add(out.pop()); 28 } 29 30 return ret; 31 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/PostorderTraversal.java
LeetCode: Binary Tree Postorder Traversal 解题报告
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原文地址：http://www.cnblogs.com/yuzhangcmu/p/4172783.html
