05 August 2008
Given two binary trees, write a function to check if they are the same or not.
Two binary trees are considered the same if they are structurally identical and the nodes have the same value.
Example 1:
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Input: 1 1
/ \ / \
2 3 2 3
[1,2,3], [1,2,3]
Output: true
Example 2:
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Input: 1 1
/ \
2 2
[1,2], [1,null,2]
Output: false
Example 3:
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Input: 1 1
/ \ / \
2 1 1 2
[1,2,1], [1,1,2]
Output: false
用递归和非递归分别实现。
递归
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
} else if (p == null || q == null) {
return false;
}
if (p.val != q.val) {
return false;
}
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
迭代的办法，可以用BFS - Queue或者DFS - Stack来逐个检查，这里用BFS-Queue
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/**
* 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 isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
}
if (p == null || q == null) {
return false;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(p);
queue.offer(q);
while (!queue.isEmpty()) {
int currentSize = queue.size();
for (int i = 0; i < currentSize; i++) {
TreeNode nodeP = queue.poll();
TreeNode nodeQ = queue.poll();
if (nodeP == null && nodeQ == null) {
continue;
}
if (nodeP == null || nodeQ == null || nodeP.val != nodeQ.val) {
return false;
}
queue.offer(nodeP.left);
queue.offer(nodeQ.left);
queue.offer(nodeP.right);
queue.offer(nodeQ.right);
}
}
return true;
}
}