LeetCode: Binary Tree Postorder Traversal 解题报告

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

 public List<Integer> postorderTraversal1(TreeNode root) {
         List<Integer> ret = new ArrayList<Integer>();
         dfs(root, ret);
         return ret;
     }

     // Solution 1: rec
     public void dfs(TreeNode root, List<Integer> ret) {
         if (root == null) {
             return;
         }

         dfs(root.left, ret);
         dfs(root.right, ret);
         ret.add(root.val);
     }

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
     */

 // Solution 2: iterator
     public List<Integer> postorderTraversal(TreeNode root) {
         List<Integer> ret = new ArrayList<Integer>();
         if (root == null) {
             return ret;
         }

         Stack<TreeNode> s = new Stack<TreeNode>();
         Stack<Integer> out = new Stack<Integer>();

         s.push(root);

         while (!s.isEmpty()) {
             TreeNode cur = s.pop();
             out.push(cur.val);

             if (cur.left != null) {
                 s.push(cur.left);
             }

             if (cur.right != null) {
                 s.push(cur.right);
             }
         }

         while (!out.isEmpty()) {
             ret.add(out.pop());
         }

         return ret;
     }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/PostorderTraversal.java