145. 二叉树的后序遍历 递归 迭代 Morris
- 二叉树的后序遍历
给你一棵二叉树的根节点 root ,返回其节点值的 后序遍历 。
示例 1:
输入:root = [1,null,2,3]
输出:[3,2,1]
示例 2:
输入:root = []
输出:[]
示例 3:
输入:root = [1]
输出:[1]
提示:
树中节点的数目在范围 [0, 100] 内
-100 <= Node.val <= 100
进阶:递归算法很简单,你可以通过迭代算法完成吗?
递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ans=new ArrayList<Integer>();
if(root==null)return ans;
ans.addAll(postorderTraversal(root.left));
ans.addAll(postorderTraversal(root.right));
ans.add(root.val);
return ans;
}
}
迭代
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<Integer>();
if (root == null) return ans;
Deque<TreeNode> stk = new LinkedList<TreeNode>();
TreeNode pre = null;
while (root != null || !stk.isEmpty()) {
while (root != null) {
stk.push(root);
root = root.left;
}
root = stk.pop();
if (root.right == null || root.right == pre) {
//第二次访问,或者是右子树为空的情况,记录答案,记录pre,置root为空
ans.add(root.val);
pre = root;
root = null;
} else {
//第一次访问节点,将自己压回栈,然后进入右子树
stk.push(root);
root = root.right;
}
}
return ans;
}
}
Morris
class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
TreeNode rt=root;
List<Integer> ans = new ArrayList<Integer>();
if (root == null) return ans;
while (root != null) {
if (root.left != null) {
TreeNode pre = root.left;
while (pre.right != null && pre.right != root) pre=pre.right;
if (pre.right == null) {
pre.right = root;
root = root.left;
} else {
pre.right = null;//这条不能省略,因为后面访问斜的路径还要以它来判断终点
addPath(ans, root.left);
root = root.right;
}
}
else {
root = root.right;
}
}
addPath(ans, rt);
return ans;
}
访问一斜行,类似于adidas
public void addPath(List<Integer> ans, TreeNode node) {
int cnt = 0;
while (node != null) {
++cnt;
ans.add(node.val);
node = node.right;
}
int l = ans.size() - cnt, r = ans.size() - 1;
while (l < r) {
int t = ans.get(l);
ans.set(l, ans.get(r));
ans.set(r, t);
l++;
r--;
}
}
// public void addPath(List<Integer> ans, TreeNode node) {
// if(node==null)return ;
// addPath(ans,node.right);
// ans.add(node.val);
// }
}