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成都住建局官网住建智慧建管,什么建站程序最利于seo,wordpress域名重复,百度商桥代码怎么加到网站上专栏:Java数据结构秘籍 个人主页:手握风云 目录 一、非递归实现遍历二叉树 1.1. 二叉树的前序遍历 1.2. 二叉树的中序遍历 1.3. 二叉树的后序遍历 一、非递归实现遍历二叉树 1.1. 二叉树的前序遍历 我们这里要使用栈来进行实现。我们反向思考一下为…

专栏:Java数据结构秘籍

个人主页:手握风云

目录

一、非递归实现遍历二叉树

1.1. 二叉树的前序遍历

1.2. 二叉树的中序遍历

1.3. 二叉树的后序遍历


一、非递归实现遍历二叉树

1.1. 二叉树的前序遍历

        我们这里要使用栈来进行实现。我们反向思考一下为什么不使用队列?如下图,前序遍历肯定是先将根结点放进去,如果是队列,根结点先进先出,然后怎么去遍历右子树呢,就无法打印的顺序了。

        我们定义一个引用cur,只要cur不为null,就打印值并将该元素放入栈中。当遍历到4时,左子树为空,返回结点4并弹出,再去遍历4的右结点,然后返回结点2并弹出,让cur等于结点2的右子树并遍历。只要1的左子树没有遍历完,1就不弹出。

public class Solution {public void preorderTraversal(TreeNode root){if(root == null){return;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while(cur != null){stack.push(cur);System.out.print(cur.val+" ");cur = cur.left;}}
}

        代码写到这里就会出现问题,原因是:当遍历到结点4的时候,4的左子树为空,就无法进入while循环。然后把4弹出去,让cur=top,问题又来了,如果结点4左边要是不为空,又得放入栈中,也需要走while循环。

        我们会发现当cur走到某个结点时,如果为空,但栈不为空,此时就可以巧妙地在while外面再加一层while循环。

while (cur != null || !stack.isEmpty()) {while (cur != null) {stack.push(cur);System.out.print(cur.val + " ");cur = cur.left;}cur = stack.pop();cur = cur.right;
}

        完整代码实现:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;class TreeNode{int val;TreeNode left;TreeNode right;public TreeNode() {}public TreeNode(int val) {this.val = val;}public TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;}
}public class Solution {public List<Integer> preorderTraversal(TreeNode root){List<Integer> tree = new ArrayList<>();if(root == null){return tree;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()) {while (cur != null) {tree.add(cur.val);stack.push(cur);cur = cur.left;}cur = stack.pop();cur = cur.right;}return tree;}
}
import java.util.ArrayList;
import java.util.List;public class Test {public static void main(String[] args) {List<Integer> result = new ArrayList<>();Solution solution = new Solution();TreeNode root = new TreeNode(1,new TreeNode(2),new TreeNode(3));root.left.left = new TreeNode(4);root.left.right = new TreeNode(5);root.left.right.left = new TreeNode(6);root.left.right.right = new TreeNode(7);root.right.right = new TreeNode(8);root.right.right.left = new TreeNode(9);result = solution.preorderTraversal(root);System.out.println(result);}
}

1.2. 二叉树的中序遍历

        与前序遍历的思路相同,只是打印的时机不一样。中序遍历要在弹出的元素之后直接打印。

        完整代码实现:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;class TreeNode{int val;TreeNode left;TreeNode right;public TreeNode() {}public TreeNode(int val) {this.val = val;}public TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;}
}public class Solution {public List<Integer> inorderTraversal(TreeNode root){List<Integer> tree = new ArrayList<>();if(root == null){return tree;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;while (cur != null || !stack.isEmpty()) {while (cur != null) {tree.add(cur.val);stack.push(cur);cur = cur.left;}cur = stack.pop();tree.add(cur.val);cur = cur.right;}return tree;}
}
import java.util.ArrayList;
import java.util.List;public class Test {public static void main(String[] args) {List<Integer> result = new ArrayList<>();Solution solution = new Solution();TreeNode root = new TreeNode(1,new TreeNode(2),new TreeNode(3));root.left.left = new TreeNode(4);root.left.right = new TreeNode(5);root.left.right.left = new TreeNode(6);root.left.right.right = new TreeNode(7);root.right.right = new TreeNode(8);root.right.right.left = new TreeNode(9);result = solution.inorderTraversal(root);System.out.println(result);}
}

1.3. 二叉树的后序遍历

        后序遍历不能按照我们上面前序与中序的方法来做。如果结点下面还有孩子结点,如果把4弹出之后,就无法获取它的右侧,所以只能获取不能弹出。当右子树为空,才能弹出,再进行打印。

public class Solution {public void postorderTraversal(TreeNode root){if(root == null){return;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;TreeNode top = null;while(cur != null || !stack.isEmpty()) {while(cur != null){stack.push(cur);cur = cur.left;}top = stack.peek();if(top.right == null){stack.pop();System.out.print(top.val+" ");}else{cur = top.right;}}}
}public class Test {public static void main(String[] args) {Solution solution = new Solution();TreeNode root = new TreeNode(1,new TreeNode(2),new TreeNode(3));root.left.left = new TreeNode(4);root.left.right = new TreeNode(5);root.left.right.right = new TreeNode(7);root.right.right = new TreeNode(8);root.right.right.left = new TreeNode(9);solution.postorderTraversal(root);}
}

        但这样写,会存在问题:当遍历到结点5的右结点7时,会陷入死循环。那我们怎么知道这个结点被打印过?我们再定义引用prev,让prev来记录被弹出的结点。

        while(cur != null || !stack.isEmpty()) {while(cur != null){stack.push(cur);cur = cur.left;}top = stack.peek();if(top.right == null || top.right == prev){stack.pop();System.out.print(top.val+" ");prev = top;}else{cur = top.right;}

        完整代码实现:

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;class TreeNode{int val;TreeNode left;TreeNode right;public TreeNode() {}public TreeNode(int val) {this.val = val;}public TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;}
}public class Solution {public List<Integer> postorderTraversal(TreeNode root){List<Integer> tree = new ArrayList<>();if(root == null){return tree;}Stack<TreeNode> stack = new Stack<>();TreeNode cur = root;TreeNode top = null;TreeNode prev = null;while(cur != null || !stack.isEmpty()) {while(cur != null){stack.push(cur);cur = cur.left;}top = stack.peek();if(top.right == null || top.right == prev){tree.add(top.val);stack.pop();prev = top;}else{cur = top.right;}}return tree;}
}
import java.util.ArrayList;
import java.util.List;public class Test {public static void main(String[] args) {List<Integer> tree = new ArrayList<>();Solution solution = new Solution();TreeNode root = new TreeNode(1,new TreeNode(2),new TreeNode(3));root.left.left = new TreeNode(4);root.left.right = new TreeNode(5);root.left.right.left = new TreeNode(6);root.left.right.right = new TreeNode(7);root.right.right = new TreeNode(8);root.right.right.left = new TreeNode(9);tree = solution.postorderTraversal(root);System.out.println(tree);}
}
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