leetcode101：对称二叉树

Question

leetcode101：对称二叉树

sisterAn opened this issue 4 years ago · comments

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。

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

进阶：

你可以运用递归和迭代两种方法解决这个问题吗？

附赠leetcode地址：leetcode

Eden · Answer 1 · Wed May 27 2020 09:37:01 GMT+0800 (China Standard Time)

递归

比较二叉树的 前序遍历 和 对称前序遍历 来判断是不是对称的。

前序遍历：根-左-右

对称前序遍历：根-右-左

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
const isSymmetric = (root) => {
  return isSymmetrical(root, root)
};

const isSymmetrical = (r1, r2) => {
  if(r1 === null && r2 === null) return true

  if(r1 === null || r2 === null) return false

  if(r1.val !== r2.val) return false

  return isSymmetrical(r1.left, r2.right) && isSymmetrical(r1.right, r2.left)
}

7777sea · Answer 2 · Wed May 27 2020 09:39:44 GMT+0800 (China Standard Time)

/**
* Definition for a binary tree node.
* function TreeNode(val) {
*     this.val = val;
*     this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isSymmetric = function(root) {

if(root === null) return true;

function isEqual(left, right){

if(left === null && right === null) return true;
if(left === null || right === null) return false;
if(left.val !== right.val) return false;

return isEqual(left.left, right.right) && isEqual(left.right, right.left)
}

return isEqual(root.left, root.right)
};

fantc · Answer 3 · Wed May 27 2020 10:02:51 GMT+0800 (China Standard Time)

递归

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if(root === null) return true;
    function isEqual(left, right){
       if(left === null && right === null) return true;
       if(left === null || right === null) return false;
       if(left.val !== right.val) return false;
       return isEqual(left.left, right.right) && isEqual(left.right, right.left)
    }
    return isEqual(root.left, root.right)
}

迭代

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if(root === null) return true;
    const list = []
    list.push([root.left, root.right])
    while(list.length>0){
        const track = list.pop()
        if(track[0] !== null && track[1] !== null){
            if(track[0].val !== track[1].val) return false;
            list.push([track[0].left, track[1].right])
            list.push([track[0].right, track[1].left])
        }else if((track[0] !== null && track[1] === null) || (track[0] === null && track[1] !== null)) {
            return false;
        }
    }
    return true
}

Eden · Answer 4 · Wed May 27 2020 10:27:19 GMT+0800 (China Standard Time)

递归

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if(root === null) return true;
    function isEqual(left, right){
       if(left === null && right === null) return true;
       if(left === null || right === null) return false;
       if(left.val !== right.val) return false;
       return isEqual(left.left, right.right) && isEqual(left.right, right.left)
    }
    return isEqual(root.left, root.right)
}

迭代

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
    if(root === null) return true;
    const list = []
    list.push([root.left, root.right])
    while(list.length>0){
        const track = list.pop()
        if(track[0] !== null && track[1] !== null){
            if(track[0].val !== track[1].val) return false;
            list.push([track[0].left, track[1].right])
            list.push([track[0].right, track[1].left])
        }else if((track[0] !== null && track[1] === null) || (track[0] === null && track[1] !== null)) {
            return false;
        }
    }
    return true
}

@fanefanes 这个迭代学到了，感谢~

jasonting5 · Answer 5 · Wed May 27 2020 22:06:59 GMT+0800 (China Standard Time)

var isSymmetric = function(root) {
    // 用迭代方法实现
    if (!root) {
        return true
    }
    let queue = []
    queue.push(root)
    queue.push(root)
    while (queue.length > 0) {
        let p1 = queue.shift(),
            p2 = queue.shift()
        if (!p1 && !p2) continue
        if (!p1 || !p2) return false
        if (p1.val !== p2.val) return false
        queue.push(p1.left)
        queue.push(p2.right)
        queue.push(p1.right)
        queue.push(p2.left)
    }
    return true
}

瓶子君 · Answer 6 · Wed May 27 2020 23:40:00 GMT+0800 (China Standard Time)

解答：

一棵二叉树对称，则需要满足：根的左右子树是镜像对称的

也就是说，每棵树的左子树都和另外一颗树的右子树镜像对称，左子树的根节点值与右子树的根节点值相等

所以，我们需要比较：

左右子树的根节点值是否相等
左右子树是否镜像对称

边界条件：

左右子树都为 null 时，返回 true
左右子树有一个 null 时，返回 false

解法一：递归

const isSymmetric = function(root) {
    if(!root) return true
    var isEqual = function(left, right) {
        if(!left && !right) return true
        if(!left || !right) return false
        return left.val === right.val
         && isEqual(left.left, right.right)
         && isEqual(left.right, right.left)
    }
    return isEqual(root.left, root.right)
};

复杂度分析：

时间复杂度：O(n)
空间复杂度：O(n)

解法二：迭代

利用栈来记录比较的过程，实际上，递归就使用了调用栈，所以这里我们可以使用栈来模拟递归的过程

首先根的左右子树入栈
将左右子树出栈，比较两个数是否互为镜像
如果左右子树的根节点值相等，则将左子树的 left 、右子树的 right 、左子树的 right 、右子树的 left 依次入栈
继续出栈（一次出栈两个进行比较）…….

依次循环出栈入栈，直到栈为空

const isSymmetric = function(root) {
    if(!root) return true
    let stack = [root.left, root.right]
    while(stack.length) {
        let right = stack.pop()
        let left = stack.pop()
        if(left && right) {
            if(left.val !== right.val) return false
            stack.push(left.left)
            stack.push(right.right)
            stack.push(left.right)
            stack.push(right.left)
        } else if(left || right) {
            return false
        }
    }
    return true
};

复杂度分析：

时间复杂度：O(n)
空间复杂度：O(n)

leetcode

芦苇 · Answer 7 · Thu May 28 2020 00:17:53 GMT+0800 (China Standard Time)

var isSymmetric = function (root) {
  if (root === null) return true;

  function symmetric(left, right) {
    if (left === null && right === null) {
      return true;
    } else if (left !== null && right !== null) {
      return (
        left.val === right.val &&
        symmetric(left.left, right.right) &&
        symmetric(left.right, right.left)
      );
    } else {
      return false;
    }
  }
  return symmetric(root.left, root.right);
};

Anran · Answer 8 · Fri Nov 20 2020 15:25:17 GMT+0800 (China Standard Time)

var isSymmetric = function(root) {
    if(!root) return true;
    let quene = [root];
    let tmpQueue = [];

    while(quene.length) {
        let len = quene.length;
        for(let i = 0;i<len;i++) {
            let node = quene.shift();
            if(node.left) {
                quene.push(node.left);
                tmpQueue.push(node.left.val);
            } else {
                tmpQueue.push('#');
            }
            if(node.right) {
                quene.push(node.right);
                tmpQueue.push(node.right.val);
            } else {
                tmpQueue.push('#');
            }
        }
        let left = 0, right = tmpQueue.length - 1;
        while(left < right) {
            if(tmpQueue[left++] !== tmpQueue[right--]) return false;
        }
        tmpQueue = [];
    }
    return true;
};

xiaowuge007 · Answer 9 · Tue Apr 27 2021 11:10:18 GMT+0800 (China Standard Time)

function symmetryTree_2(root) {
	if(!root){return false}
	if(!root.left && !root.right){return true}
	if(!root.left || !root.right){return false}
	if(root.left && root.right){
		let arr = [root.left,root.right];
		while (arr.length>0){
			let left = arr.shift();
			let right = arr.pop();
			if(!left && !right){
				continue;
			}
			if((left && !right) || (right && !right) || (left.id !== right.id)){
				return false
			}
			arr.unshift(left.left);
			arr.push(right.right);
			arr.unshift(left.right);
			arr.push(right.left);
		}
		return true
	}
}