【每日一题Day110】LC2331计算布尔二叉树的值 | dfs

计算布尔二叉树的值【LC2331】

You are given the root of a full binary tree with the following properties:

• Leaf nodes have either the value 0 or 1, where 0 represents False and 1 represents True.
• Non-leaf nodes have either the value 2 or 3, where 2 represents the boolean OR and 3 represents the boolean AND.

The evaluation of a node is as follows:

• If the node is a leaf node, the evaluation is the value of the node, i.e. True or False.
• Otherwise, evaluate the node’s two children and apply the boolean operation of its value with the children’s evaluations.

Return the boolean result of evaluating the root node.

A full binary tree is a binary tree where each node has either 0 or 2 children.

A leaf node is a node that has zero children.

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• 思路：DFS

  使用深度搜索，返回每颗子树的布尔值，每颗子树的布尔结果与其布尔操作有关，即父节点的值，因此递归函数需要传入父节点的值

• 实现

  • 确定递归函数的参数和返回值

    • 参数：当前节点root，父节点的值

    • 返回值：boolean，该子树的计算结果

  • 确定终止条件

    • 当前节点是空时返回，当父节点为或操作时，返回false；当父节点为与操作时，返回true；

  • 确定单层递归的逻辑

    • 当节点是叶子节点时，根据值返回true和false
    • 当节点不是叶子节点使，继续dfs搜索

  /**
   * 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 evaluateTree(TreeNode root) {
          return dfs(root, 2);
      }
      public boolean dfs (TreeNode root, int pre){
          if (root == null){
              if (pre == 2) return false;// or
              else return true;// and
          }
          if (root.val == 2) {
              return dfs(root.left, 2) || dfs(root.right, 2);
          }
          else if (root.val == 3){
              return dfs(root.left, 3) && dfs(root.right, 3);
          }
          return root.val == 1;
      }
  
  }
  

  • 复杂度
    • 时间复杂度：$O(n)$
    • 空间复杂度：$O(n)$

• 优化：不需要记录父节点的值，二叉树为完整二叉树

  class Solution {
      public boolean evaluateTree(TreeNode root) {
          if (root.left == null) {
              return root.val == 1;
          } 
          if (root.val == 2) {
              return evaluateTree(root.left) || evaluateTree(root.right);
          } else {
              return evaluateTree(root.left) && evaluateTree(root.right);
          }
      }
  }
  
  作者：力扣官方题解
  链接：https://leetcode.cn/problems/evaluate-boolean-binary-tree/solutions/2091770/ji-suan-bu-er-er-cha-shu-de-zhi-by-leetc-4g8f/
  来源：力扣（LeetCode）
  著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
  

  • 复杂度
    • 时间复杂度：$O(n)$
    • 空间复杂度：$O(n)$