【数据结构】算法 Binary Tree Postorder Traversal 二叉树后序遍历

目录

• Binary Tree Postorder Traversal 二叉树后序遍历
  • 思路
• Tag

Binary Tree Postorder Traversal 二叉树后序遍历

后序遍历是啥就不说了，百度一下， 简单说就是 left_child->right_child->root.

Input: root = [1,null,2,3]
Output: [3,2,1]

思路

递归处理

 /**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> list = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        if(root==null){
            return list;
        }

         postorderTraversal(root.left);
         postorderTraversal(root.right);
        list.add(root.val);
         return list;
    }
}

非递归的方式

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> ans =new ArrayList<>();
        Stack<Integer> s2 = new Stack<Integer>();//记录遍历方向
        Stack<TreeNode> s1 = new Stack<TreeNode>();//记录treenode轨迹
        if(root==null){
            return ans;
        }
        s1.push(root);
        s2.push(0);
        while(!s1.isEmpty()){
            root = s1.peek();
            int status = s2.pop();
            switch (status){
                case 0://左
                    s2.push(1);
                    if(root.left!=null){
                        s1.push(root.left);
                        s2.push(0);
                    }
                    break;
                case 1://右
                    s2.push(2);
                    if(root.right!=null){
                        s1.push(root.right);
                        s2.push(0);
                    }
                    break;
                case 2://root
                    ans.add(s1.peek().val);
                    s1.pop();
            }
        }
        return ans;
    }
}

Tag

stack