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