Binary Tree Postorder Traversal

佚名 6年前 (2019-04-23) 随笔 1650人围观抢沙发百度已收录

https://blog.csdn.net/dutsoft/article/details/37737725

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

For example:
Given binary tree {1,#,2,3},

return [3,2,1].

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

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

思路：二叉树的递归后序遍历，非递归的也得会。