另一棵树的子树

• https://leetcode.cn/problems/subtree-of-another-tree/description/

描述

• 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]
• -$10^4$ <= root.val <= $10^4$
• -$10^4$ <= subRoot.val <= $10^4$

Typescript 版算法实现

1 ) 方案1：深度优先搜索暴力匹配

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/function isSubtree(root: TreeNode | null, subRoot: TreeNode | null): boolean {if (!root) return false;return check(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
}function check(a: TreeNode | null, b: TreeNode | null): boolean {if (!a && !b) return true;if (!a || !b) return false;if (a.val === b.val) return check(a.left, b.left) && check(a.right, b.right);return false;
}

2 ) 方案2：深度优先搜索序列上做串匹配

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/function getMaxElement(root: TreeNode | null, maxElement: { value: number }): void {if (!root) return;maxElement.value = Math.max(maxElement.value, root.val);getMaxElement(root.left, maxElement);getMaxElement(root.right, maxElement);
}function getDfsOrder(root: TreeNode | null, order: number[], lNull: number, rNull: number): void {if (!root) return;order.push(root.val);if (root.left) {getDfsOrder(root.left, order, lNull, rNull);} else {order.push(lNull);}if (root.right) {getDfsOrder(root.right, order, lNull, rNull);} else {order.push(rNull);}
}function kmp(sOrder: number[], tOrder: number[]): boolean {const sLen = sOrder.length, tLen = tOrder.length;const fail: number[] = new Array(tLen).fill(-1);for (let i = 1, j = -1; i < tLen; ++i) {while (j !== -1 && tOrder[i] !== tOrder[j + 1]) {j = fail[j];}if (tOrder[i] === tOrder[j + 1]) {++j;}fail[i] = j;}for (let i = 0, j = -1; i < sLen; ++i) {while (j !== -1 && sOrder[i] !== tOrder[j + 1]) {j = fail[j];}if (sOrder[i] === tOrder[j + 1]) {++j;}if (j === tLen - 1) {return true;}}return false;
}function isSubtree(s: TreeNode | null, t: TreeNode | null): boolean {let maxElement = Number.MIN_SAFE_INTEGER;getMaxElement(s, { value: maxElement });getMaxElement(t, { value: maxElement });const lNull = maxElement + 1;const rNull = maxElement + 2;const sOrder: number[] = [];const tOrder: number[] = [];getDfsOrder(s, sOrder, lNull, rNull);getDfsOrder(t, tOrder, lNull, rNull);return kmp(sOrder, tOrder);
}

3 ) 方案3：基于两棵树是否相同来判断

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/function isSubtree(root:TreeNode | null, subRoot:TreeNode | null) {// 不停地比较， 某一个子树，是不是和subRoot一样if(!root) return falseif((root.val === subRoot.val) && (isSameTree(root,subRoot))) return truereturn isSubtree(root.left,subRoot) || isSubtree(root.right,subRoot)
};function isSameTree(p: TreeNode | null, q: TreeNode | null): boolean {if(!p && !q) return trueif(!p || !q) return falseif(p.val!==q.val) return falsereturn isSameTree(p.left,q.left) && isSameTree(p.right,q.right)
};