README.md

572. 另一个树的子树

English Version

题目描述

给定两个非空二叉树 s 和 t，检验 s 中是否包含和 t 具有相同结构和节点值的子树。s 的一个子树包括 s 的一个节点和这个节点的所有子孙。s 也可以看做它自身的一棵子树。

示例 1:
给定的树 s:

     3
    / \
   4   5
  / \
 1   2

给定的树 t：

   4 
  / \
 1   2

返回 true，因为 t 与 s 的一个子树拥有相同的结构和节点值。

示例 2:
给定的树 s：

     3
    / \
   4   5
  / \
 1   2
    /
   0

给定的树 t：

   4
  / \
 1   2

返回 false。

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubtree(self, root: TreeNode, subRoot: TreeNode) -> bool:
        def dfs(root1, root2):
            if root1 is None and root2 is None:
                return True
            if root1 is None or root2 is None:
                return False
            return root1.val == root2.val and dfs(root1.left, root2.left) and dfs(root1.right, root2.right)

        if root is None:
            return False
        return dfs(root, subRoot) or self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null) {
            return false;
        }
        return dfs(root, subRoot) || isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
    }

    private boolean dfs(TreeNode root1, TreeNode root2) {
        if (root1 == null && root2 == null) {
            return true;
        }
        if (root1 == null || root2 == null) {
            return false;
        }
        return root1.val == root2.val && dfs(root1.left, root2.left) && dfs(root1.right, root2.right);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if (!root) return 0;
        return dfs(root, subRoot) || isSubtree(root->left, subRoot) || isSubtree(root->right, subRoot);
    }

    bool dfs(TreeNode* root1, TreeNode* root2) {
        if (!root1 && !root2) return 1;
        if (!root1 || !root2) return 0;
        return root1->val == root2->val && dfs(root1->left, root2->left) && dfs(root1->right, root2->right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSubtree(root *TreeNode, subRoot *TreeNode) bool {
	if root == nil {
		return false
	}
	var dfs func(root1, root2 *TreeNode) bool
	dfs = func(root1, root2 *TreeNode) bool {
		if root1 == nil && root2 == nil {
			return true
		}
		if root1 == nil || root2 == nil {
			return false
		}
		return root1.Val == root2.Val && dfs(root1.Left, root2.Left) && dfs(root1.Right, root2.Right)
	}
	return dfs(root, subRoot) || isSubtree(root.Left, subRoot) || isSubtree(root.Right, subRoot)
}

JavaScript

/**
 * 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} root
 * @param {TreeNode} subRoot
 * @return {boolean}
 */
var isSubtree = function (root, subRoot) {
    if (!root) return false;
    let dfs = function (root1, root2) {
        if (!root1 && !root2) {
            return true;
        }
        if (!root1 || !root2) {
            return false;
        }
        return (
            root1.val == root2.val &&
            dfs(root1.left, root2.left) &&
            dfs(root1.right, root2.right)
        );
    };
    return (
        dfs(root, subRoot) ||
        isSubtree(root.left, subRoot) ||
        isSubtree(root.right, subRoot)
    );
};

...