剑指Offer之Java算法习题精讲二叉树专题篇下_Java教程

剑指Offer之Java算法习题精讲二叉树专题篇下

2022-10-08 15:30明天一定. Java教程

跟着思路走，之后从简单题入手，反复去看，做过之后可能会忘记，之后再做一次，记不住就反复做，反复寻求思路和规律，慢慢积累就会发现质的变化

题目一

剑指Offer之Java算法习题精讲二叉树专题篇下

解法

/**
* 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 int maxDepth(TreeNode root) {
      return  method(root);
  }
  int method(TreeNode root){
      if(root==null){
          return 0;
      }
      int l = method(root.left);
      int r = method(root.right);
      return Math.max(l, r) + 1;
  }
}

题目二

剑指Offer之Java算法习题精讲二叉树专题篇下

解法

/**
* 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 {
  int ans = 0;
  public int diameterOfBinaryTree(TreeNode root) {
      method(root);
      return ans;
  }
  public int method(TreeNode root){
      if(root==null){
          return 0;
      }
      int l = method(root.left);
      int r = method(root.right);
      ans = Math.max(ans,l+r);
      return Math.max(l,r)+1;
  }
}

题目三

剑指Offer之Java算法习题精讲二叉树专题篇下

解法

/**
* 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 int minDepth(TreeNode root) {
      if(root==null) return 0;
      if(root.left==null&&root.right==null) return 1;
      int min = Integer.MAX_VALUE;
      if(root.left!=null){
          min = Math.min(min,minDepth(root.left));
      }
      if(root.right!=null){
          min = Math.min(min,minDepth(root.right));
      }
      return min+1;        
  }
}

题目四

剑指Offer之Java算法习题精讲二叉树专题篇下

解法

/**
* 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<Integer>();
  public List<Integer> preorderTraversal(TreeNode root) {
      method(root);
      return list;
  }
  public void method(TreeNode root){
      if(root==null){
          return;
      }
      // 前序
      list.add(root.val);
      method(root.left);
      // 中序
      method(root.right);
      // 后序
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123510927