LeetCode 654.最大二叉树

题目讲解

思路

二叉树的根是最大值
左子树最大值左边部分构造的最大二叉树
右子树最大值是右边部分构造的最大二叉树

在这里插入图片描述

AC代码

/**
 * 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 constructMaximumBinaryTree(int[] nums) {
        return constructMaximumBinaryTree1(nums,0,nums.length);

    }
    public TreeNode constructMaximumBinaryTree1(int[]nums, int leftIndex,int rightIndex)
    {
        if( rightIndex-leftIndex <1)
        {
            return null;
        }
        if( rightIndex-leftIndex==1)
        {
            return  new TreeNode(nums[leftIndex]);
        }
        int maxIndex=leftIndex;
        int maxVal= nums[maxIndex];
        for(int i=leftIndex+1;i<rightIndex;i++)
        {
            if( nums[i]>maxVal)
            {
                maxVal =nums[i];
                maxIndex =i;
            }
        }
       TreeNode root= new TreeNode ( maxVal);
       root.left= constructMaximumBinaryTree1(nums,leftIndex,maxIndex);
       root.right =constructMaximumBinaryTree1(nums,maxIndex+1,rightIndex);
       return root;
    }
}

LeetCode 617.合并二叉树

题目讲解

难点

如何同时操作两个二叉树

代码实现简单到不行
在这里插入图片描述

/**
 * 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 mergeTrees(TreeNode root1, TreeNode root2) {
      if( root1 ==null)
      {
          return root2;
      }
      if( root2==null)
      {
          return root1;
      } 
      root1.val += root2.val;
      //左边
      root1.left = mergeTrees( root1.left,root2.left);
      //右边
      root1.right =mergeTrees(root1.right,root2.right);
      return root1;
    }
}

Leetcode 700.二叉搜索树中的搜索

题目讲解

思路

二叉搜索树是一个有序树：

若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值；
若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
它的左、右子树也分别为二叉搜索树

/**
 * 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 searchBST(TreeNode root, int val) {
      //运用搜索树的特点
      if( root == null || root.val ==val)
      {
          return root;
      }
      if( val<root.val )
      {
          return searchBST(root.left,val);
      }
      else {

      return searchBST(root.right,val);
    }
    }
}

Leetcode 98.验证二叉搜索树

题目讲解

思路

遇到搜索树要记得用上中序遍历可以验证
在结束条件那块 if( root ==null) return 回去的应该是 boolean类型的true 因为
null 也是二叉搜索树

AC代码

/**
 * 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 {
    //递归
    TreeNode max;
    public boolean isValidBST(TreeNode root) {
     if(root==null)
     {
         return true; //null也是二叉树
     }
     //  左
     boolean left = isValidBST(root.left);
     if(!left )
     {
         return false;
     }
     //中
     if( max !=null && root.val <= max.val )
     {
         return false;
     }
     max =root;
   
    //右边
    boolean right =isValidBST(root.right);
    return right;


    }
}