100. Same Tree - jiejackyzhang/leetcode-note GitHub Wiki

Given two binary trees, write a function to check if they are equal or not.

Two binary trees are considered equal if they are structurally identical and the nodes have the same value.

Tree类题目。

##Approach 1: recursive 典型的recursive problem。

basic cases为：

  1. p和q都为null，return true；
  2. p和q只有一个为null，return false；
  3. p和q都不为null，且p.val与q.val不等，return false。

否则，进行递归，返回p，q左右两个子树结果的与。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if(p == null && q == null) return true;
        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);
    }
}

##Approach 2: iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(p);
        queue.offer(q);
        while(!queue.isEmpty()) {
            TreeNode node1 = queue.poll();
            TreeNode node2 = queue.poll();
            if(node1 == null && node2 == null) continue;
            if(node1 == null || node2 == null) return false;
            if(node1.val != node2.val) return false;
            queue.offer(node1.left);
            queue.offer(node2.left);
            queue.offer(node1.right);
            queue.offer(node2.right);
        }
        return true;
    }
}
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