给你两棵二叉树 root 和 subRoot 。检验 root 中是否包含和 subRoot 具有相同结构和节点值的子树。如果存在,返回 true ;否则,返回 false 。
二叉树 tree 的一棵子树包括 tree 的某个节点和这个节点的所有后代节点。tree 也可以看做它自身的一棵子树。
示例 1:
输入:root = [3,4,5,1,2], subRoot = [4,1,2]
输出:true
示例 2:
输入:root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
输出:false
提示:
root 树上的节点数量范围是 [1, 2000]
subRoot 树上的节点数量范围是 [1, 1000]
-104 <= root.val <= 104
-104 <= subRoot.val <= 104
通过次数115,674提交次数243,636
请问您在哪类招聘中遇到此题?
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/subtree-of-another-tree
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//递归算法 class Solution { public: bool isSubtree(TreeNode* root, TreeNode* subRoot) { if(!root&&!subRoot)return true; if(!root||!subRoot)return false; if(isSameTree(root,subRoot))return true; bool a = isSubtree(root->left,subRoot); bool b = isSubtree(root->right,subRoot); bool c = (a||b); return c; } bool isSameTree(TreeNode*root,TreeNode*subRoot) { if(!root&&!subRoot)return true; if(!root||!subRoot)return false; if(root->val!=subRoot->val)return false; bool a = isSameTree(root->left,subRoot->left); bool b = isSameTree(root->right,subRoot->right); bool c = (a&&b); return c; } }; //使用队列 class Solution { public: bool isSubtree(TreeNode* root, TreeNode* subRoot) { queue<TreeNode*>qu; qu.push(root); while(!qu.empty()) { TreeNode* Node = qu.front(); qu.pop(); bool a = isSameTree(Node,subRoot); if(a) { return true; } if(Node->left)qu.push(Node->left); if(Node->right)qu.push(Node->right); } return false; } bool isSameTree(TreeNode*root,TreeNode*subRoot) { queue<TreeNode*>qu1; qu1.push(root); qu1.push(subRoot); while(!qu1.empty()) { TreeNode* Node1 = qu1.front(); qu1.pop(); TreeNode* Node2 = qu1.front(); qu1.pop(); if(!Node1&&!Node2)continue; if(!Node1||!Node2||(Node1->val!=Node2->val))return false; qu1.push(Node1->left); qu1.push(Node2->left); qu1.push(Node1->right); qu1.push(Node2->right); } return true; } }; //使用stack class Solution { public: bool isSubtree(TreeNode* root, TreeNode* subRoot) { stack<TreeNode*>qu; qu.push(root); while(!qu.empty()) { TreeNode* Node = qu.top(); qu.pop(); bool a = isSameTree(Node,subRoot); if(a) { return true; } if(Node->left)qu.push(Node->left); if(Node->right)qu.push(Node->right); } return false; } bool isSameTree(TreeNode*root,TreeNode*subRoot) { stack<TreeNode*>qu1; qu1.push(root); qu1.push(subRoot); while(!qu1.empty()) { TreeNode* Node1 = qu1.top(); qu1.pop(); TreeNode* Node2 = qu1.top(); qu1.pop(); if(!Node1&&!Node2)continue; if(!Node1||!Node2||(Node1->val!=Node2->val))return false; qu1.push(Node1->left); qu1.push(Node2->left); qu1.push(Node1->right); qu1.push(Node2->right); } return true; } };