[leetcode]572. Subtree of Another Tree

本文介绍了一种通过先序遍历并使用StringBuilder存储树结构的方法来判断一棵树是否为另一棵树的子树。通过实例展示了如何实现该算法。

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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

思路:先序遍历,将树存到StringBuilder里,

Solution:

package com.billkang;

class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;

	TreeNode(int x) {
		val = x;
	}
}

/**
 * @author binkang
 * @date May 21, 2017
 */
public class SubtreeofAnotherTree {
	public boolean isSubtree(TreeNode s, TreeNode t) {
		StringBuilder sbd1 = new StringBuilder();
		StringBuilder sbd2 = new StringBuilder();

		preOrder(sbd1, s);
		preOrder(sbd2, t);
		return sbd1.toString().contains(sbd2);
	}

	private void preOrder(StringBuilder sbd, TreeNode t) {
		if(t == null) {
			sbd.append(",,");
			return;
		}
		
		sbd.append(",").append(t.val);
		preOrder(sbd, t.left);
		preOrder(sbd, t.right);
	}
}
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