145. 二叉樹的後序遍歷
知識點：二叉樹；遞迴；Morris遍歷
題目描述
給定一個二叉樹的根節點 root ，返回它的 後序 遍歷。
示例
輸入: [1,null,2,3]
1
\
2
/
3
輸出: [3,2,1]
解法一：遞迴
/**
* 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 {
List<Integer> list = new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) {
if(root == null) return list;
postorderTraversal(root.left);
postorderTraversal(root.right);
list.add(root.val);
return list;
}
}
時間複雜度：0(N),每個節點恰好被遍歷一次；
空間複雜度：O(N),遞迴過程中棧的開銷；
解法二:迭代法
後序遍歷可以用前序遍歷來解決，想一下前序遍歷：根左右，我們先壓右樹再壓左樹。怎麼實現根右左呢，可以先壓左樹再壓右樹嘛，然後反過來不就是左右根了嗎？（反過來用棧來實現，棧一個很大的作用就是實現逆序）
/**
* 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 {
List<Integer> list = new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) {
Stack<TreeNode> stackA = new Stack<>();
Stack<TreeNode> stackB = new Stack<>();
if(root == null) return list;
stackA.push(root);
while(!stackA.isEmpty()){
TreeNode top = stackA.pop();
stackB.push(top);
if(top.left != null) stackA.push(top.left);
if(top.right != null) stackA.push(top.right);
}
while(!stackB.isEmpty()){
list.add(stackB.pop().val);
}
return list;
}
}
解法三：Morris遍歷
構建從下到上的連線，一條路能夠走遍所有節點；
當我們返回上層之後，也就是將連線斷開的時候，列印下層的單鏈表。
/**
* 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 {
List<Integer> list = new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) {
if(root == null) return list;
TreeNode cur = root;
TreeNode mostRightNode = null;
while(cur != null){
mostRightNode = cur.left;
if(mostRightNode != null){
while(mostRightNode.right != null && mostRightNode.right != cur){
mostRightNode = mostRightNode.right;
}
if(mostRightNode.right == null){
mostRightNode.right = cur;
cur = cur.left;
continue;
}else{
mostRightNode.right = null;
postMorrisPrint(cur.left); //第二次到達時列印下一層的單鏈表；
cur = cur.right;
}
}else{
cur = cur.right;
}
}
postMorrisPrint(root);
return list;
}
private void postMorrisPrint(TreeNode node){
TreeNode reverseList = reverseList(node); //反轉單鏈表；
TreeNode cur = reverseList;
while(cur != null){
list.add(cur.val);
cur = cur.right;
}
reverseList(reverseList);
}
private TreeNode reverseList(TreeNode node){
TreeNode pre = null;
TreeNode cur = node;
while(cur != null){
TreeNode next = cur.right;
cur.right = pre;
pre = cur;
cur = next;
}
return pre;
}
}
體會
後序遍歷的特殊在於其Morris遍歷；