145. 二叉树的后序遍历

LeetCode_145

题目描述

递归解法

/**
 * 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 {
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        inTravels(root);
        return list;
    }
    public void inTravels(TreeNode now){
        if(now == null)
            return;
        inTravels(now.left);
        inTravels(now.right);
        list.add(now.val);
    }
}

非递归解法

再来看后序遍历，先序遍历是中左右，后续遍历是左右中，那么我们只需要调整一下先序遍历的代码顺序，就变成中右左的遍历顺序，然后在反转result数组，输出的结果顺序就是左右中了，如下图：

/**
 * 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 {
    List<Integer> list = new ArrayList<>();
    public List<Integer> postorderTraversal(TreeNode root) {
        Stack<TreeNode> sta = new Stack<>();
        if(root == null)
            return list;
        sta.push(root);
        while(!sta.isEmpty()){
            TreeNode now = sta.pop();
            list.add(now.val);
            if(now.left != null)
                sta.push(now.left);
            if(now.right != null)
                sta.push(now.right);
        }
        Collections.reverse(list);
        return list;
    }
}