LeetCode 二叉树后序遍历（binary-tree-postorder-traversal）

 

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

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

   1
    \
     2
    /
   3

 

return[3,2,1].

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

 

import java.util.*;
public class Solution {
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        if(root==null)
            return list;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        Stack<TreeNode> stack2 = new Stack<TreeNode>();
        stack.add(root);
        while(!stack.isEmpty()){
            TreeNode r = stack.pop();
            if(r.left!=null)
                stack.add(r.left);
            if(r.right!=null)
                stack.add(r.right);
            stack2.add(r);
        }
        while(!stack2.isEmpty()){
            list.add(stack2.pop().val);
        }
        //LRD(list,root);递归
        return list;
    }
    public void LRD(ArrayList<Integer> list,TreeNode root){
        if(root==null)
            return;
        LRD(list,root.left);
        LRD(list,root.right);
        list.add(root.val);
    }
}

 

转载于:https://www.cnblogs.com/googlemeoften/p/5819044.html