[leetcode]Binary Tree Postorder Traversal

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?

算法思路：

最简单的后序遍历。不解释了。

代码如下：

public class Solution {
    List<Integer> res = new ArrayList<Integer>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root == null) return res;
        if(root.left != null) postorderTraversal(root.left);
        if(root.right != null) postorderTraversal(root.right);
        res.add(root.val);
        return res;
    }
}

 

非递归实现：

需要两个栈，一个数字栈用来记录后序遍历的逆序，一个节点栈来遍历整棵树，如果不明白也没关系，有个绝招！背下来吧。。。。

代码如下：

public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        if(root == null) return res;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<Integer> intStack = new Stack<Integer>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            intStack.push(node.val);
            if(node.left != null) stack.push(node.left);
            if(node.right != null) stack.push(node.right);
        }
        while(!intStack.isEmpty()){
            res.add(intStack.pop());
        }
        return res;
    }
}