leetcode-100

Given the root nodes p and q of two binary trees, write a function to check whether the two trees are the same.
Two trees are considered identical if they are structurally identical and the nodes have the same value.

Tip:
The number of nodes on both trees is in the range [0, 100]
-104 <= Node.val <= 104

Node definition (java version):

/**
 * 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;
 *     }
 * }
 */

Idea 1:
Using the idea of ​​depth-first traversal, two trees are traversed, and the leaf nodes are compared one by one during the traversal process.

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // 判断为null的情况，两个节点都为null，则返回true，不再遍历
        if(p == null && q == null){
            return true;
        }
        // 一个节点为null，则返回false，不再遍历
        if(p == null){
            return false;
        }
        if(q == null){
            return false;
        }
        // 如果两个节点都不为null，且两个节点的值不一致，则返回false，不再遍历
        if(p.val != q.val){
            return false;
        }
        // 递归遍历两个节点的左节点是否一致，若左节点一致，则再遍历右节点
        if(isSameTree(p.left,q.left)){
            return isSameTree(p.right, q.right);
        }
        return false;
    }
}

Idea 2:
Using the idea of ​​breadth-first traversal, two trees are traversed, and the leaf nodes are compared one by one during the traversal process.
Breadth-first traversal uses the first-in-first-out feature of the queue to flatten the tree and compare them one by one

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        Queue<TreeNode> queueLeft = new LinkedList();
        Queue<TreeNode> queueRight = new LinkedList();
        queueLeft.add(p);
        queueRight.add(q);
        while(!queueLeft.isEmpty() && !queueRight.isEmpty()){
            TreeNode l = queueLeft.poll();
            TreeNode r = queueRight.poll();
            if(l == null && r != null ){
                return false;
            }
            // 注意：两个对比的节点都为null的时候，说明这两个节点相等，还需要排查后续节点，所以是continue
            if(l == null && r == null){
                continue;
            }
            if(l != null && r == null){
                return false;
            }
            if(l.val == r.val){
                queueLeft.add(l.left);
                queueLeft.add(l.right);
                queueRight.add(r.left);
                queueRight.add(r.right);
                continue;
            }
            return false;
        }
        return queueLeft.isEmpty() && queueRight.isEmpty();
    }
}