872. 叶子相似的树-蒲公英云

872. 叶子相似的树

你的名字 2023-10-05 21:09 108阅读 0赞

请考虑一棵二叉树上所有的叶子，这些叶子的值按从左到右的顺序排列形成一个 叶值序列 。

举个例子，如上图所示，给定一棵叶值序列为 (6, 7, 4, 9, 8) 的树。

如果有两棵二叉树的叶值序列是相同，那么我们就认为它们是 叶相似 的。

如果给定的两个头结点分别为 root1 和 root2 的树是叶相似的，则返回 true；否则返回 false 。

示例 1：

输入：root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
输出：true

示例 2：

输入：root1 = [1], root2 = [1]
输出：true

示例 3：

输入：root1 = [1], root2 = [2]
输出：false

示例 4：

输入：root1 = [1,2], root2 = [2,2]
输出：true

示例 5：

输入：root1 = [1,2,3], root2 = [1,3,2]
输出：false

提示：

给定的两棵树可能会有 1 到 200 个结点。

给定的两棵树上的值介于 0 到 200 之间。

package Solution872;

import java.util.Stack;

public class Solution {

public boolean leafSimilar(TreeNode root1, TreeNode root2) {
    // Create empty stacks. These stacks are going
    // to be used for iterative traversals.
    Stack<TreeNode> s1 = new Stack<TreeNode>();
    Stack<TreeNode> s2 = new Stack<TreeNode>();
    s1.push(root1);
    s2.push(root2);
    // Loop until either of two stacks is not empty
    while (!s1.empty() || !s2.empty()) {
        // If one of the stacks is empty means other
        // stack has extra leaves so return false
        if (s1.empty() || s2.empty())
            return false;
        TreeNode temp1 = s1.pop();
        while (temp1 != null && !(temp1.left == null && temp1.right == null)) {
            // Push right and left children of temp1.
            // Note that right child is inserted
            // before left
            if (temp1.right != null)
                s1.push(temp1.right);
            if (temp1.left != null)
                s1.push(temp1.left);
            temp1 = s1.pop();
        }
        // same for tree2
        TreeNode temp2 = s2.pop();
        while (temp2 != null && !(temp2.left == null && temp2.right == null)) {
            if (temp2.right != null)
                s2.push(temp2.right);
            if (temp2.left != null)
                s2.push(temp2.left);
            temp2 = s2.pop();
        }
        // If one is null and other is not, then
        // return false
        if (temp1 == null && temp2 != null)
            return false;
        if (temp1 != null && temp2 == null)
            return false;
        // If both are not null and data is not
        // same return false
        if (temp1 != null && temp2 != null) {
            if (temp1.val != temp2.val)
                return false;
        }
    }
    // If control reaches this point, all leaves
    // are matched
    return true;
}
static void inorder(TreeNode node) {
    if (node == null)
        return;
    /* first recur on left child */
    inorder(node.left);
    /* then print the data of node */
    System.out.printf("%d ", node.val);
    /* now recur on right child */
    inorder(node.right);
}
public static void main(String[] args) {
    // Let us create trees in above example 1
    Solution sol = new Solution();
    TreeNode root1 = new TreeNode(1);
    root1.left = new TreeNode(2);
    root1.right = new TreeNode(3);
    root1.left.left = new TreeNode(4);
    root1.right.left = new TreeNode(6);
    root1.right.right = new TreeNode(7);
    TreeNode root2 = new TreeNode(0);
    root2.left = new TreeNode(1);
    root2.right = new TreeNode(5);
    root2.left.right = new TreeNode(4);
    root2.right.left = new TreeNode(6);
    root2.right.right = new TreeNode(7);
    if (sol.leafSimilar(root1, root2))
        System.out.println("Same");
    else
        System.out.println("Not Same");
}