代码随想录算法训练营第二十天|654 最大二叉树 617 合并二叉树 700 二叉搜索树中的搜索 98 验证二叉搜索树

654 最大二叉树

class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return buildHelper(nums,0,nums.length);
    }
    private TreeNode buildHelper(int[] nums,int start,int end){
        if(end == start)return null;
        int mid = start;
        for(int i = start;i < end;i++){
            if(nums[mid] < nums[i]){
                mid = i;
            }
        }
        TreeNode cur = new TreeNode(nums[mid]);
        cur.left = buildHelper(nums,start,mid);
        cur.right = buildHelper(nums,mid + 1,end);
        return cur;
    }
}

时间复杂度O(n^2)递归所需要的时间×遍历nums所需要的时间

空间复杂度O(n)需要使用的栈空间

617 合并二叉树 

递归 

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null)return root2;
        if(root2 == null)return root1;
        root1.val += root2.val;
        root1.left = mergeTrees(root1.left,root2.left);
        root1.right = mergeTrees(root1.right,root2.right);
        return root1;
    }
}

时间复杂度O(min(n,m))n,m为两棵树中节点的个数

空间复杂度O(min(n,m))

迭代

class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if(root1 == null)return root2;
        if(root2 == null)return root1;
        Dequest = new LinkedList<>();
        st.push(root2);
        st.push(root1);
        while(!st.isEmpty()){
            TreeNode node1 = st.pop();
            TreeNode node2 = st.pop();
            node1.val += node2.val;
            if(node1.left != null && node2.left != null){
                st.push(node2.left);
                st.push(node1.left);
            }else if(node1.left == null)node1.left = node2.left;
            if(node1.right != null && node2.right != null){
                st.push(node2.right);
                st.push(node1.right);
            }else if(node1.right == null)node1.right = node2.right;
        }
        return root1;
    }
}

时间复杂度O(min(n,m))

空间复杂度O(min(n,m))

700 二叉搜索树中的搜索

 二叉搜索树：在val小的在根节点左边，val大的在根节点右边

迭代

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

时间复杂度O(n)

空间复杂度O(n)

递归

class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if(root == null)return null;
        if(root.val == val)return root;
        return root.val < val ? searchBST(root.right,val) : searchBST(root.left,val);
    }
}

时间复杂度O(n)

空间复杂度O(1)

98 验证二叉搜索树