LeetCode题解-145-Binary Tree Postorder Traversal

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原题

解法概览

解法1为递归；

解法2位迭代。

解法1

解题思路

递归法。后序遍历左子树，后序遍历右子树，访问根节点。

代码

public class Solution145_recursive {    public List<Integer> postorderTraversal(TreeNode root) {        ArrayList<Integer> treeValList = new ArrayList<Integer>();        if (root == null)            return treeValList;        ArrayList<Integer> leftTreeValList = (ArrayList<Integer>) postorderTraversal(root.left);        ArrayList<Integer> rightTreeValList = (ArrayList<Integer>) postorderTraversal(root.right);        treeValList.addAll(leftTreeValList);        treeValList.addAll(rightTreeValList);        treeValList.add(root.val);        return treeValList;    }}

解法2

解题思路

1)当栈顶指针非空，应遍历左子树（遍历时不访问节点）；

2)若从左子树返回，如果当前节点的右子树为空或者已被访问，那么右子树进栈，否则出栈，访问该节点。

图解

看不清图的话可以在浏览器 页面中单独打开。

代码

<pre name="code" class="java">public class Solution145_iterator {    public List<Integer> postorderTraversal(TreeNode root) {        Stack<TreeNode> treeNodeStack = new Stack<TreeNode>();        ArrayList<Integer> treeNodeList = new ArrayList<Integer>();        treeNodeStack.push(root);        TreeNode lastVisit = null;        while (!treeNodeStack.empty()){            while (treeNodeStack.peek() != null){                TreeNode currentNode = treeNodeStack.peek();                treeNodeStack.push(currentNode.left);            }            treeNodeStack.pop();            if (!treeNodeStack.isEmpty()){                TreeNode currentNode = treeNodeStack.peek();                if (currentNode.right == null || currentNode.right == lastVisit){                    lastVisit = treeNodeStack.pop();                    treeNodeList.add(lastVisit.val);                    treeNodeStack.push(null);                }                else{                    treeNodeStack.push(currentNode.right);                }            }        }        return treeNodeList;    }}