剑指Offer之Java算法习题精讲二叉树专题篇上

来和二叉树玩耍吧~

题目一

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 boolean isSymmetric(TreeNode root) {
      return method(root.left,root.right);
  }
  public boolean method(TreeNode l,TreeNode r){
      if(l==null&&r==null) return true;
      if(l==null||r==null||l.val!=r.val) return false;
      return method(l.left,r.right)&&method(l.right,r.left);
  }
}

题目二

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 TreeNode sortedArrayToBST(int[] nums) {
       return method(nums,0,nums.length-1);
  }
  public TreeNode method(int[] nums,int l,int r){
      if(l>r) return null;
      int mid = l+(r-l)/2;
      TreeNode root = new TreeNode(nums[mid]);
      root.left = method(nums,l,mid-1);
      root.right = method(nums,mid+1,r);
      return root;
  }
}

题目三

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 boolean isBalanced(TreeNode root) {
      if(root==null) return true;
      return Math.abs(method(root.left)-method(root.right))<=1&&isBalanced(root.left)&&isBalanced(root.right);
  }
  public int method(TreeNode root){
      if(root==null) return 0;
      return Math.max(method(root.left),method(root.right))+1;
  }
}

题目四

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 boolean hasPathSum(TreeNode root, int targetSum) {
      if(root==null) return false;
      if(root.left == null && root.right == null) return targetSum==root.val;
      return hasPathSum(root.left,targetSum-root.val)||hasPathSum(root.right,targetSum-root.val);
  }
}

题目五

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 TreeNode invertTree(TreeNode root) {
      if(root==null) return null;
      TreeNode node = new TreeNode(root.val);
      node.right = invertTree(root.left);
      node.left = invertTree(root.right);
      return node;
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123532571