目录

1,题目

2，代码

2.1深度优先遍历

2.2广度优先遍历

3，学习与总结

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1,题目

给你两棵二叉树的根节点 p 和 q ，编写一个函数来检验这两棵树是否相同。

如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。

2，代码

2.1深度优先遍历

如果两个二叉树都为空，则二叉树相同；

如果两个二叉树只有一个为空，则两个二叉树一定不相同；

如果两个二叉树都不为空，则先判断根节点的值是否相同；若相同，则分别判断两个二叉树的左子树和右子树是否相同；

/*** Definition for a binary tree node.* function TreeNode(val, left, right) {*     this.val = (val===undefined ? 0 : val)*     this.left = (left===undefined ? null : left)*     this.right = (right===undefined ? null : right)* }*/
/*** @param {TreeNode} p* @param {TreeNode} q* @return {boolean}*/
var isSameTree = function(p, q) {// 深度优先搜索 递归  树结构是否一样 再去比较节点中的值是否是一样的if(p === null && q ===null){return true;}else if (p===null || q===null){return false;}else if(p.val != q.val  ){return false;}else {// 该节点 的子树结构一样 且值一样，则继续递归return isSameTree(p.left,q.left) && isSameTree(p.right,q.right)}
};

2.2广度优先遍历

锻炼思维

/*** Definition for a binary tree node.* function TreeNode(val, left, right) {*     this.val = (val===undefined ? 0 : val)*     this.left = (left===undefined ? null : left)*     this.right = (right===undefined ? null : right)* }*/
/*** @param {TreeNode} p* @param {TreeNode} q* @return {boolean}*/
var isSameTree = function(p, q) {if (p === null && q === null) {return true;} else if (p === null || q === null) {return false;}let queue1 = [p];let queue2 = [q];while (queue1.length > 0 && queue2.length > 0) {let node1 = queue1.shift();let node2 = queue2.shift();if (node1.val !== node2.val) {return false;}let left1 = node1.left, right1 = node1.right, left2 = node2.left, right2 = node2.right;if ((left1 === null) !== (left2 === null)) {return false;}if ((right1 === null) !== (right2 === null)) {return false;}if (left1 !== null) {queue1.push(left1);}if (right1 !== null) {queue1.push(right1);}if (left2 !== null) {queue2.push(left2);}if (right2 !== null) {queue2.push(right2);}}return queue1.length === 0 && queue2.length === 0;
};

3，学习与总结

广度优先思维需要继续加强