Binary Tree Postorder Traversal:迭代后续遍历二叉树

Given a binary tree, return the postorder traversal of its nodes' values.

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

思路一：后序遍历：左，右，根。所以对于每个节点，用三种颜色-1,0,1表示左节点未访问，左节点已访问但右节点未访问，左右节点均已访问。对于均已访问的节点，则保存结果。

思路二：以根、右、左的方式先序遍历，然后将结果逆序，就变成了左、右、根的顺序。更简洁。

思路代码1：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans = new ArrayList<Integer>();
        if(root == null ) return ans;
        Stack<TreeNode> stackNode = new Stack<TreeNode>();
        Stack<Integer> stackFlag = new Stack<Integer>();
        stackNode.push(root);
        stackFlag.push(-1);
        while(!stackNode.isEmpty()){
            TreeNode top = stackNode.peek();
            Integer flag = stackFlag.peek();
            if(top == null){
                stackNode.pop();
                stackFlag.pop();
                continue;
            }
            if(flag == -1){
                stackFlag.pop();
                stackFlag.push(0);
                if(stackNode.peek().left != null){
                    stackNode.push(stackNode.peek().left);
                    stackFlag.push(-1);
                }    
            }else if(flag == 0){
                stackFlag.pop();
                stackFlag.push(1);
                stackNode.push(stackNode.peek().right);
                stackFlag.push(-1);             
            }else{
                ans.add(stackNode.pop().val);
                stackFlag.pop();
            }           
        }
        return ans;
    }
}

思路2代码：

class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new LinkedList<>();
    	Deque<TreeNode> stack = new ArrayDeque<>();
    	while (root != null || !stack.isEmpty()) {
    		if (root != null) {
    			stack.push(root);
    			res.add(0, root.val);
    			root = root.right;
    		} else {
    			root = stack.pop();
    			root = root.left;
    		}
    	}
    	return res;
    }
}