代码随想录 Day14 - 二叉树（一）

2023-07-13
江苏
本文字数：1541 字
阅读完需：约 5 分钟

作业题

递归法遍历二叉树

144. 二叉树的前序遍历

/** * 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> preorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<>();        dfs(root, result);        return result;    }        private void dfs(TreeNode node, List<Integer> result) {        if (node == null) {            return;        }         result.add(node.val);        dfs(node.left, result);        dfs(node.right, result);    }}

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94. 二叉树的中序遍历

/** * 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> inorderTraversal(TreeNode root) {        List<Integer> result = new ArrayList<>();        dfs(root, result);        return result;    }        private void dfs(TreeNode node, List<Integer> result) {        if (node == null) {            return;        }        dfs(node.left, result);        result.add(node.val);        dfs(node.right, result);    }}

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145. 二叉树的后序遍历

/** * 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> result = new ArrayList<>();        dfs(root, result);        return result;    }        private void dfs(TreeNode node, List<Integer> result) {        if (node == null) {            return;        }        dfs(node.left, result);        dfs(node.right, result);        result.add(node.val);    }}