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面试题 26. 树的子结构

题目描述

输入两棵二叉树 A 和 B，判断 B 是不是 A W 的子结构。(约定空树不是任意一个树的子结构)

B 是 A 的子结构， 即 A 中有出现和 B 相同的结构和节点值。

例如:

给定的树 A:

     3
    / \
   4   5
  / \
 1   2

给定的树 B：

   4 
  /
 1

返回 true，因为 B 与 A 的一个子树拥有相同的结构和节点值。

示例 1：

输入：A = [1,2,3], B = [3,1]
输出：false

示例 2：

输入：A = [3,4,5,1,2], B = [4,1]
输出：true

限制：

• 0 <= 节点个数 <= 10000

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isSubStructure(self, A: TreeNode, B: TreeNode) -> bool:
        def dfs(A, B):
            if B is None:
                return True
            if A is None or A.val != B.val:
                return False
            return dfs(A.left, B.left) and dfs(A.right, B.right)
        
        if A is None or B is None:
            return False
        return dfs(A, B) or self.isSubStructure(A.left, B) or self.isSubStructure(A.right, B)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubStructure(TreeNode A, TreeNode B) {
        if (A == null || B == null) {
            return false;
        }
        return dfs(A, B) || isSubStructure(A.left, B) || isSubStructure(A.right, B);
    }

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

JavaScript

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

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isSubStructure(A *TreeNode, B *TreeNode) bool {
	var dfs func(A, B *TreeNode) bool
	dfs = func(A, B *TreeNode) bool {
		if B == nil {
			return true
		}
		if A == nil || A.Val != B.Val {
			return false
		}
		return dfs(A.Left, B.Left) && dfs(A.Right, B.Right)
	}
	if A == nil || B == nil {
		return false
	}
	return dfs(A, B) || isSubStructure(A.Left, B) || isSubStructure(A.Right, B)
}

C++

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

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

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