572. Subtree of Another Tree（子树）

发布时间：2020-12-14 04:36:26 所属栏目：大数据来源：网络整理

导读：Given two non-empty binary trees?s?and?t,check whether tree?t?has exactly the same structure and node values with a subtree of?s. A subtree of?s?is a tree consists of a node in?sand all of this node‘s descendants. The tree?s?could also be

Given two non-empty binary trees?s?and?t,check whether tree?t?has exactly the same structure and node values with a subtree of?s. A subtree of?s?is a tree consists of a node in?sand all of this node‘s descendants. The tree?s?could also be considered as a subtree of itself.

Example 1:
Given tree s:

     3
    /    4   5
  /  1   2

Given tree t:

   4 
  /  1   2

Return?true,because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

Given tree t:

   4
  /  1   2
题目描述：判断t是不是s的子树

方法一：递归
时间复杂度：o（n）             空间复杂度：o（1）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSubtree(TreeNode s,TreeNode t) {
        if(s==null) return false;
　　　　　// 考虑几种情况，第一种s=t，调用isSubtreeStartRoot(s,t)返回true。第二种情况在根的左子树或右子树上，这需要递归了，因为不知道具体从哪个点开始。
        return isSubtreeStartRoot(s,t)||isSubtree(s.left,t)||isSubtree(s.right,t); 
    }
    private boolean isSubtreeStartRoot(TreeNode s,TreeNode t){
        if(s==null&&t==null) return true;  //两棵树都走完了，说明两棵树一样，返回true
        if(s==null||t==null) return false;  //一颗树走完了，另一颗没走完，那说明这两个不一样，返回false。
        if(s.val!=t.val) return false;
        return isSubtreeStartRoot(s.left,t.left)&&isSubtreeStartRoot(s.right,t.right);
    }
}

（编辑：李大同）

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