572. Subtree of Another Tree (E)

Subtree of Another Tree (E)

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 s and 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:

     3
    / \
   4   5
  / \
 1   2
    /
   0

Given tree t:

   4
  / \
 1   2

Return false.

---

题意

判断一个树t是否与另一个树s的子树相同。

思路

1. dfs处理。递归遍历s，对每个结点判断以该结点为根的子树是否与t相同。
2. 官方解答中的另一种方法：对两树进行前序遍历，得到类似"#1lnullrnull"的字符串ss和tt，判断tt是否是ss的子串。

---

代码实现 - dfs

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        // t一定不为null
        if (s == null) {
            return false;
        }
        return judge(s, t) || isSubtree(s.left, t) || isSubtree(s.right, t);
    }

    private boolean judge(TreeNode s, TreeNode t) {
        if (s == null && t == null) {
            return true;
        } else if (s == null || t == null || s.val != t.val) {
            return false;
        }
        return judge(s.left, t.left) && judge(s.right, t.right);
    }
}

代码处理 - 前序遍历

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
        String ss = preOrder(s, false);
        String tt = preOrder(t, false);
        return ss.indexOf(tt) != -1;
    }

    private String preOrder(TreeNode root, boolean isLeft) {
        if (root == null) {
            return isLeft ? "lnull" : "rnull";
        }
        return "#" + root.val + preOrder(root.left, true) + preOrder(root.right, false);
    }
}