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LeetCode 二叉树反序遍历(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); }}
LeetCode 二叉树反序遍历(binary-tree-postorder-traversal)
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