标签:pac iss turn continue || var space treenode else
题目描述: 给定一个二叉树,检查它是否是镜像对称的。
例如,二叉树 [1,2,2,3,4,4,3] 是对称的。
DFS:递归,和100题相同的树类似,不过要注意是左右子树进行比较
//C //注意这个函数声明 bool isMirroTree(struct TreeNode* p, struct TreeNode* q); bool isSymmetric(struct TreeNode* root){ return isMirroTree(root, root); } bool isMirroTree(struct TreeNode* p, struct TreeNode* q){ if(p == NULL && q == NULL) return true; else if(p == NULL || q == NULL) return false; else if(p -> val == q -> val){ return isMirroTree(p -> left,q -> right) && isMirroTree(p -> right, q -> left); } return false; } //JS var isSymmetric = function(root) { let isMirroTree = function(p, q){ if(!p && !q) return true; if(!p || !q) return false; if(p.val != q.val) return false; return isMirroTree(p.left, q.right) && isMirroTree(p.right, q.left); }; return isMirroTree(root, root); };
BFS:迭代,用队列层次遍历,可以使用双队列,也可以单队列
//C //双队列实现 #define QUEUESIZE 200 bool isSymmetric(struct TreeNode* root){ if(!root) return 1; if(!root->left && !root->right) return 1; if(!root->left || !root->right) return 0; /*init queue*/ struct TreeNode* q1[QUEUESIZE]; struct TreeNode* q2[QUEUESIZE]; int front1, rear1, front2, rear2; front1 = rear1 = front2 = rear2 = -1; q1[++rear1] = root->left; q2[++rear2] = root->right; struct TreeNode *a, *b; while(front1 < rear1){ a = q1[++front1]; b = q2[++front2]; if(!a && !b) continue; if(!a || !b) return 0; if(a->val != b->val) return 0; q1[++rear1] = a->left; q2[++rear2] = b->right; q1[++rear1] = a->right; q2[++rear2] = b->left; } return 1; } //单队列实现 #define QUEUESIZE 200 bool isSymmetric(struct TreeNode* root){ if(!root) return 1; if(!root->left && !root->right) return 1; if(!root->left || !root->right) return 0; /*init queue*/ struct TreeNode* q[QUEUESIZE]; int front, rear; front = rear = -1; q[++rear] = root; q[++rear] = root; struct TreeNode *a, *b; while(front < rear){ a = q[++front1]; b = q[++front2]; if(!a && !b) continue; if(!a || !b) return 0; if(a->val != b->val) return 0; q[++rear] = a->left; q[++rear] = b->right; q[++rear] = a->right; q[++rear] = b->left; } return 1; } //JS //双队列 var isSymmetric = function(root) { if(!root) return true; let leftTreeQueue = [], rightTreeQueue = [], leftNode, rightNode; leftTreeQueue.push(root.left); rightTreeQueue.push(root.right); while(leftTreeQueue.length != 0 && rightTreeQueue.length != 0){ leftNode = leftTreeQueue.shift(); rightNode = rightTreeQueue.shift(); if(!leftNode && !rightNode) continue; if(!leftNode || !rightNode) return false; if(leftNode.val != rightNode.val) return false; else{ leftTreeQueue.push(leftNode.left); rightTreeQueue.push(rightNode.right); leftTreeQueue.push(leftNode.right); rightTreeQueue.push(rightNode.left); } } return true; }; //单队列 var isSymmetric = function(root) { if(!root) return true; let treeQueue = [], leftNode, rightNode; treeQueue.push(root); treeQueue.push(root); while(treeQueue.length != 0){ leftNode = treeQueue.shift(); rightNode = treeQueue.shift(); if(!leftNode && !rightNode) continue; if(!leftNode || !rightNode) return false; if(leftNode.val != rightNode.val) return false; else{ treeQueue.push(leftNode.left); treeQueue.push(rightNode.right); treeQueue.push(leftNode.right); treeQueue.push(rightNode.left); } } return true; };
标签:pac iss turn continue || var space treenode else
原文地址:https://www.cnblogs.com/JesseyWang/p/13111707.html