力扣——二叉树的后序遍历
给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3] 1 \ 2 / 3 输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
/** * 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) {//非递归写法 List<Integer> res = new ArrayList<Integer>(); if(root == null) return res; Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode pre = null; stack.push(root); while(!stack.isEmpty()){ TreeNode curr = stack.peek(); if((curr.left == null && curr.right == null) || (pre != null && (pre == curr.left || pre == curr.right))){ //如果当前结点左右子节点为空或上一个访问的结点为当前结点的子节点时,当前结点出栈 res.add(curr.val); pre = curr; stack.pop(); }else{ if(curr.right != null) stack.push(curr.right); //先将右结点压栈 if(curr.left != null) stack.push(curr.left); //再将左结点入栈 } } return res; } }