Java数据结构第十六期：走进二叉树的奇妙世界(五)

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

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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); } }