250. Count Univalue Subtrees

题目:

Given a binary tree, count the number of uni-value subtrees.

A Uni-value subtree means all nodes of the subtree have the same value.

For example:
Given binary tree,

              5
             / 
            1   5
           /    
          5   5   5

return 4.

链接: http://leetcode.com/problems/count-univalue-subtrees/

题解:

求Uni-value subtree的个数。对树来说这求count的首先思路就是递归了,不过这里要另外构造一个辅助函数来判断root为顶点的subtree是否是Uni-value subtree,假如是Uni-value的subtree,则结果可以+1,否则,我们返回递归求出的左子树的count和右节点的count。假如是Python的话可以一边计算一边返回。Java写起来可能稍微麻烦点。

Time Complexity - O(n), Space Complexity - O(n).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        if(root.left == null && root.right == null)
            return 1;
        if(root.left == null) {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.right) + 1;
            else
                return countUnivalSubtrees(root.right);
        } else if (root.right == null) {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.left) + 1;
            else
                return countUnivalSubtrees(root.left);
        } else {
            if(isUniValueSubtree(root))
                return countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right) + 1;
            else
                return countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right);
        }
    }
    
    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;
        if(root.left == null && root.right == null)
            return true;
        else if (root.left != null && root.right != null) {
            if(root.val == root.left.val && root.val == root.right.val)
                return isUniValueSubtree(root.left) && isUniValueSubtree(root.right);
            else
                return false;
        } else if (root.left == null) {
            if(root.right.val == root.val)
                return isUniValueSubtree(root.right);
            else
                return false;
        } else {
            if(root.left.val == root.val)
                return isUniValueSubtree(root.left);
            else
                return false;
        }
    }
}

Update: 优化一下

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        if(root.left == null && root.right == null)
            return 1;
        int res = countUnivalSubtrees(root.left) + countUnivalSubtrees(root.right) ;
        return isUniValueSubtree(root) ? res + 1 : res;
    }
    
    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;
        if(root.left == null && root.right == null)
            return true;
        if(isUniValueSubtree(root.left) && isUniValueSubtree(root.right)) {
            if(root.left != null && root.right != null)
                return (root.left.val == root.right.val) && (root.left.val == root.val);
            else if(root.left != null)
                return root.left.val == root.val;
            else
                return root.right.val == root.val;
        }   
        return false;
    }
}

Update: 再优化一下。注意判断两个子树是否都为Uni-value subtree的时候用了 '&',这样才能判断两个条件,否则第一个条件为false时就不判断第二个了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int count = 0;
    public int countUnivalSubtrees(TreeNode root) {
        if(root == null)
            return 0;
        isUniValueSubtree(root);
        return count;
    }
    
    private boolean isUniValueSubtree(TreeNode root) {
        if(root == null)
            return true;
       
        if(isUniValueSubtree(root.left) & isUniValueSubtree(root.right)) {
            if(root.left != null && root.left.val != root.val)
                return false;
            if(root.right != null && root.right.val != root.val)
                return false;
            count++;
            return true;
        }   
        return false;
    }
}

题外话:

这道题题号是250,马克一下。

二刷:

一刷写得比较250,现在要尝试不那么250。要做这道题我们首先要看清楚例子,以及弄明白什么是subtree。 subtree就是指,在一个tree里的某一个node,以及这个node的所有descendants.那么从题目给的例子里,我们复合条件的subtree有4个, 左边第三层里的两个"5"算其2, 右边第二层的"5 - 5",以及第三层的"5"也都算符合条件的subtree,所以我们返回4。 要注意root节点及右边第二层和第三层一起组成的"5-5-5"并不是subtree,这只能算一条root-to-leaf path。另外我们再看一个例子[1, 2, 3],这里作为leaf节点的2和3也分别都是符合条件的subtree,我们返回2。其实所有的leaf节点都是符合条件的。所以我们这道题的处理方法和https://leetcode.com/problems/balanced-binary-tree/ 有点像,都是自底向上进行判断。

  1. 这里我们先创建一个Global变量count, 也可以不用global变量而使用一个size为数组作为参数传递进辅助函数isUnival中。
  2. isUnival主要就是递归地自底向上判断tree是否为unival,也就是所有节点值均相等。有下面几种情况需要考虑
    1. root = null,这时候我们返回true。递归终止时,每个leaf节点都为true,所以null情况我们返回true
    2. 当左子树为unival并且右子树也为unival的时候
      1. 假如左右子节点都为空,则我们遇到leaf节点,合理,count++
      2. 假如左子节点或者右子节点不为空,但值又不与root的值相等时,root本身加左右子树形成的这个subtree不满足题意,我们返回false
      3. 否则,root及其descendants形成一个满足题意的subtree,我们把count++,记录下这个subtree
    3. 要注意判断左右子树是否为unival的时候,左子树和右子树都要判断,这里我们使用了一个'&'符号来避免短路条件判断。也可以分开写。

Java:

Time Complexity - O(n), Space Complexity - O(n).

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int count = 0;
    
    public int countUnivalSubtrees(TreeNode root) {
        if (root == null) return 0;
        isUnival(root);
        return count;
    }
    
    private boolean isUnival(TreeNode root) {
        if (root == null) return true;
        if (isUnival(root.left) & isUnival(root.right)) {
            if (root.left != null && root.left.val != root.val) return false;
            if (root.right != null && root.right.val != root.val) return false;
            count++;
            return true;
        }
        return false;
    }
}

Reference:

https://leetcode.com/discuss/66805/my-java-solution-using-a-boolean-helper-function

https://leetcode.com/discuss/68057/very-easy-java-solution-post-order-recursion

https://leetcode.com/discuss/63286/7-line-accepted-code-in-java

https://leetcode.com/discuss/52210/c-one-pass-recursive-solution

https://leetcode.com/discuss/50241/recursive-java-solution-with-explanation

https://leetcode.com/discuss/50357/my-concise-java-solution

https://leetcode.com/discuss/50420/java-11-lines-added

https://leetcode.com/discuss/68057/very-easy-java-solution-post-order-recursion

https://github.com/google/google-java-format

https://leetcode.com/discuss/55213/my-ac-java-code

https://leetcode.com/discuss/53431/java-solution-with-dfs

https://leetcode.com/discuss/50452/ac-clean-java-solution

https://en.wikipedia.org/wiki/Tree_(data_structure) 

https://leetcode.com/problems/balanced-binary-tree/

原文地址:https://www.cnblogs.com/yrbbest/p/5011791.html