872. Leaf-Similar Trees.md

872. Leaf-Similar Trees

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leetcode-cn Daily Challenge on May 10th, 2021.

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Difficulty : Easy

Related Topics : Tree、DFS

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Consider all the leaves of a binary tree. From left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return true if and only if the two given trees with head nodes root1 and root2` are leaf-similar.

Constraints:

• Both of the given trees will have between 1 and 200 nodes.
• Both of the given trees will have values between 0 and 200.

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Solution

• Java

  • mine

    DFS Runtime: 0 ms, faster than 100.00%, Memory Usage: 37.2 MB, less than 7.14% of Java online submissions

    // O(N)Time O(N)Space
    public boolean leafSimilar(TreeNode root1, TreeNode root2) {
        List<Integer> list1 = new ArrayList<>();
        List<Integer> list2 = new ArrayList<>();
        dfs(root1,list1);
        dfs(root2,list2);
        if(list1.size() != list2.size()){
            return false;
        }
        for(int i = 0; i < list1.size(); i++){
            if(list1.get(i)!=list2.get(i)){
                return false;
            }
        }
        return true;
    }
    
    public TreeNode dfs(TreeNode node, List<Integer> list){
        if(node == null){
            return null;
        }
        TreeNode left = dfs(node.left, list);
        TreeNode right = dfs(node.right, list);
        if(left == null && right == null){
            list.add(node.val);
        }
        return node;
    }
    

  • the most votes

    **** Runtime: 0 ms, faster than 100.00%, Memory Usage: 36.7 MB, less than 89.29% of Java online submissions

    // O(N)Time O(H)Space
    public boolean leafSimilar(TreeNode root1, TreeNode root2) {
        Stack<TreeNode> s1 = new Stack<>(), s2 = new Stack<>();
        s1.push(root1); s2.push(root2);
        while (!s1.empty() && !s2.empty())
            if (dfs(s1) != dfs(s2)) return false;
        return s1.empty() && s2.empty();
    }
    
    public int dfs(Stack<TreeNode> s) {
        while (true) {
            TreeNode node = s.pop();
            if (node.right != null) s.push(node.right);
            if (node.left != null) s.push(node.left);
            if (node.left == null && node.right == null) return node.val;
        }
    }