剑指Offer之Java算法习题精讲二叉树专项解析

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 {
  int ans;
  int pre;
  public int getMinimumDifference(TreeNode root) {
      ans = Integer.MAX_VALUE;
      pre = -1;
      method(root);
      return ans;
  }
  public void method(TreeNode root){
      if(root==null) return;
      method(root.left);
      if(pre==-1){
          pre = root.val;
      }else{
          ans = Math.min(ans,root.val-pre);
          pre = root.val;
      }
      method(root.right);
  }
}

 

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 {
  int ans = 0;
  public int findTilt(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.abs(l-r);
      return l+r+root.val;
  }
}

 

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 isSubtree(TreeNode root, TreeNode subRoot) {
      return dfs(root,subRoot);
  }
  public boolean dfs(TreeNode root, TreeNode subRoot){
      if(root==null) return false;
      return cheack(root,subRoot)||dfs(root.left,subRoot)||dfs(root.right,subRoot);
  }
  public boolean cheack(TreeNode root, TreeNode subRoot){
      if(root==null&&subRoot==null) return true;
      if(root==null||subRoot==null||root.val!=subRoot.val) return false;
      return cheack(root.left,subRoot.left)&&cheack(root.right,subRoot.right);
  }
}

 

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 isSameTree(TreeNode p, TreeNode q) {
      if(p==null&&q==null) return true;
      if(p==null||q==null||q.val!=p.val) return false;
      return isSameTree(p.left,q.left)&&isSameTree(p.right,q.right);
  }
}

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原文链接：https://blog.csdn.net/wai_58934/article/details/123430512

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