leetcode(NOWCODER)---binary-tree-postorder-traversal
阿新 • • 發佈:2018-12-24
時間限制:1秒 空間限制:32768K 熱度指數:36539
本題知識點: 樹 leetcode
演算法知識視訊講解
題目描述
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?
方案一(簡單遞迴):
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ import java.util.*; public class Solution { private ArrayList<Integer> list = new ArrayList<Integer>(); public ArrayList<Integer> postorderTraversal(TreeNode root) { postOrder(root); return list; } private void postOrder(TreeNode root){ if(root != null){ postOrder(root.left); postOrder(root.right); list.add(root.val); } } }
方案二(藉助棧):
按照根、左孩子、右孩子的順序入棧,彈出根,插入到list首部,再彈出有孩子和左孩子,插入到list的首部,這樣list中的順序恰好就是後序遍歷的順序了
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
import java.util.*;
public class Solution {
private ArrayList<Integer> list = new ArrayList<Integer>();
public ArrayList<Integer> postorderTraversal(TreeNode root) {
if(root == null){
return list;
}
ArrayDeque<TreeNode> stack = new ArrayDeque<TreeNode>();
stack. push(root);
while(stack.size() > 0){
TreeNode node = stack.pop();
list.add(0, node.val);
if(node.left != null){
stack.push(node.left);
}
if(node.right != null){
stack.push(node.right);
}
}
return list;
}
}