leetcode 404,530,543,563,572,589,617,637,700

404

 　　按理说也可以递归做。

    public static int sumOfLeftLeaves(TreeNode root) {
        int total = 0;
        LinkedList<TreeNode> stack = new LinkedList<>();
        stack.push(root);
        while (!stack.isEmpty()){
            TreeNode pop = stack.pop();
            TreeNode left = pop.left;
            if (left != null){
                if (left.left == null && left.right == null){
                    total+=left.val;
                }
                pop.left = null;
                stack.push(left);
            }
            TreeNode right = pop.right;
            if (right != null){
                pop.right = null;
                stack.push(right);
            }
        }

        return total;
    }

530 

 　　比较low的做法是像我这样

class Solution {
    public int getMinimumDifference(TreeNode root) {
        List<Integer> list = new ArrayList<>();
        find(root,list);
        int min = Integer.MAX_VALUE;
        for (int i = 1; i < list.size(); i++) {
            min = Math.min(list.get(i)-list.get(i-1),min);
        }
        return min;
    }
    private  void find(TreeNode node,List<Integer> list){
        if (node.left != null){
            find(node.left,list);
        }
        list.add(node.val);
        if (node.right != null){
            find(node.right,list);
        }

    }
}

　　这儿看了下官解是用了临时变量来做  效率会好很多

class Solution {
    int min=100000000,p,q=-100000000;
    public int getMinimumDifference(TreeNode root) {
        if(root==null){
            return 0;
        }
        getMinimumDifference(root.left);
        p = root.val;
        if((p-q)<min){
            min = p-q;
        }
        q = p;
        getMinimumDifference(root.right);
        return min;
    }
}

作者：pu-ti-shu-xia-n
链接：https://leetcode-cn.com/problems/minimum-absolute-difference-in-bst/solution/zhe-yao-bao-li-ni-que-ding-bu-yong-ma-by-n002/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

543

class Solution {
      int max;
    public int diameterOfBinaryTree(TreeNode root) {
        if(root == null){
            return 0;
        }
        max = Integer.MIN_VALUE;
       deep(root,0);
       return max;
    }
    private  int deep(TreeNode node,int current){
        if (node == null){
            return current-1;
        }
        int ld = deep(node.left, current + 1);
        int rd = deep(node.right, current + 1);
        max = Math.max((ld-current)+(rd-current),max);
        return Math.max(ld,rd);
    }
}

559

class Solution {
    int max;
    public int maxDepth(Node root) {
        max = 0;
        if(root != null){
            handle(root,1);
        }
       
        return max;
    }

    public void handle(Node root,int level){
        max = Math.max(max,level);
        if(root.children != null){
            for(Node n : root.children){
                handle(n,level+1);
            }
        }
    }
}

563 树的坡度

class Solution {
     int total;
    public int findTilt(TreeNode root) {
        if(root == null){
            return 0;
        }
        total = 0;
        deep(root);
        return total;
    }

    private   int deep(TreeNode node){
        if (node == null){
            return 0;
        }

        int deepL = deep(node.left);
        int deepR = deep(node.right);
        total+= Math.abs(deepR-deepL);
        return deepL+deepR+node.val;
    }
}

572 子树

class Solution {
    public boolean isSubtree(TreeNode s, TreeNode t) {
               if (s == null || t == null){
            return false;
        }
        return handle(s,t);
    }
    private static boolean handle(TreeNode s, TreeNode t){
        if (s.val == t.val&&check(s,t)){
            return true;
        }
        if (s.left != null && handle(s.left,t)){
            return true;
        }
        if (s.right != null && handle(s.right,t)){
            return true;
        }
        return false;
    }

    private static boolean check(TreeNode s, TreeNode t){
        if (s == null && t == null){
            return true;
        }
        if ((s==null && t != null) || (s!=null && t == null)){
            return false;
        }
        if (s.val != t.val){
            return false;
        }
        return check(s.left,t.left)&&check(s.right,t.right);
    }
}

589

class Solution {
    public List<Integer> preorder(Node root) {
             List<Integer> result = new ArrayList<>();
        if (root != null) fill(root,result);
         return result;
    }
        private void fill(Node root,List<Integer> result){
        result.add(root.val);
        if (root.children!=null){
            for (Node child : root.children) {
                fill(child,result);
            }
        }
    }
}

　　迭代也就是利用栈来保存下，贴一份官解

class Solution {
    public List<Integer> preorder(Node root) {
        LinkedList<Node> stack = new LinkedList<>();
        LinkedList<Integer> output = new LinkedList<>();
        if (root == null) {
            return output;
        }

        stack.add(root);
        while (!stack.isEmpty()) {
            Node node = stack.pollLast();
            output.add(node.val);
            Collections.reverse(node.children);
            for (Node item : node.children) {
                stack.add(item);
            }
        }
        return output;
    }
}

作者：LeetCode
链接：https://leetcode-cn.com/problems/n-ary-tree-preorder-traversal/solution/ncha-shu-de-qian-xu-bian-li-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

617 合并二叉树

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
 TreeNode result = new TreeNode();
        if (root1 == null || root2 == null){
            result = root1 == null?root2:root1;
        }else {
            meage(result,root1,root2);
        }
        return result;
    }
public static void meage(TreeNode root,TreeNode root1,TreeNode root2){
        root.val = root1.val+root2.val;
        if (root1.left != null && root2.left != null){
            TreeNode lefs = new TreeNode();
            root.left = lefs;
            meage(lefs,root1.left,root2.left);
        }else {
            root.left = root1.left == null?root2.left:root1.left;
        }
        if (root1.right != null && root2.right != null){
            TreeNode lefs = new TreeNode();
            root.right = lefs;
            meage(lefs,root1.right,root2.right);
        }else {
            root.right = root1.right == null?root2.right:root1.right;
        }
    }
}

637 此题用数组存数据应该效率能提升很多

 int max;
    Map<Integer,List<Integer>> recoreds;
    public List<Double> averageOfLevels(TreeNode root) {
        List<Double> result = new ArrayList<>();
        if (root == null)return result;
        max = 0;
        recoreds = new HashMap<>();
        find(root,1);
        for (int i = 1; i <= max; i++) {
            //double asDouble = recoreds.get(i).stream().mapToDouble(Number::doubleValue).average().getAsDouble();
            int total = 0;
            List<Integer> integers = recoreds.get(i);
            for (Integer integer : integers) {
                total+=integer;
            }
            result.add((double)(total/integers.size()));
        }
        return result;
    }

    public void find(TreeNode node,int level){
        if (node == null)return;
        max = Math.max(level,max);
        List<Integer> rec = recoreds.get(level);
        if (rec == null){
            rec = new ArrayList<>();
            recoreds.put(level,rec);
        }
        rec.add(node.val);
        find(node.left,level+1);
        find(node.right,level+1);
    }

653 两数之和

class Solution {
Set<Integer> set;
    boolean exist = false;
    int s;
    public boolean findTarget(TreeNode root, int k) {
        s = k;exist = false;set = new HashSet<>();
        find(root);
        return exist;
    }

    public void find(TreeNode root){
        if (exist || root == null){
            return ;
        }
        if (set.contains(s-root.val)){
            exist = true;
            return;
        }else {
            set.add(root.val);
            find(root.left);
            find(root.right);
        }


    }
}

700 二叉树搜索

class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root != null ){
            if (root.val == val){
                return root;
            }else if (root.val>val){
                return searchBST(root.left,val);
            }else {
                return searchBST(root.right,val);
            }
        }
        return null;
    }
}