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Here's a tree for you Complete binary tree The root of the , This tree has the following characteristics ：

• Leaf node Or the value is 0 Or the value is 1 , among 0 Express False ,1 Express True .
• Nonleaf node Or the value is 2 Or the value is 3 , among 2 Represent logical or OR ,3 Representation logic and AND .

Calculation The value of a node is as follows ：

• If the node is a leaf node , Then the node value For itself , namely True perhaps False .
• otherwise , Calculation Node value of two children , Then, the operator of this node is used to compare two child values operation .

Return root node root Boolean operation value of .

Complete binary tree Each node has 0 A or 2 A child's binary tree .

Leaf node It's a node without children .

Example 1：

 Input ：root = [2,1,3,null,null,0,1]
 Output ：true
 explain ： The figure above shows the calculation process .
AND  The value of and operation node is  False AND True = False .
OR  The value of the operation node is  True OR False = True .
 The value of the root node is  True , So we go back to  true .

Example 2：

 Input ：root = [0]
 Output ：false
 explain ： The root node is the leaf node , And the value is  false, So we go back to  false .

Problem solving

Method 1 ： After the sequence traversal

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    bool dfs(TreeNode* root){
    
        if(root->val==0) return false;
        if(root->val==1) return true;
        bool left=dfs(root->left);
        bool right=dfs(root->right);
        if(root->val==2){
    
            return left||right;
        }
        else if(root->val==3){
    
            return left&&right;
        }
        return true;
        
    }
    bool evaluateTree(TreeNode* root) {
    
        return dfs(root);
    }
};