代码随想录Day19|二叉树（六）

今日任务

 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 

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654.最大二叉树

当不确定一个新的解决方案是否正确的时候

请优先使用暴力解的方式

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        return research(0, nums.length, nums);
    }
    public TreeNode research(int start, int end, int[] nums){
        if (start >= end) return null;
        int maxs = nums[start];
        int index = start;
        for(int i = start+1; i < end; i++){
            if (nums[i] > maxs){
                maxs = nums[i];
                index = i;
            }
        }
        TreeNode root = new TreeNode(maxs);
        root.left = research(start, index, nums);
        root.right = research(index + 1, end, nums);
        return root;
    }
}

617.合并二叉树

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        return search(root1, root2);
    }
    public TreeNode search(TreeNode root1, TreeNode root2){
        if(root1 == null && root2 == null) return null;
        if(root1 == null){
            return root2;
        }
        if(root2 == null){
            return root1;
        }
        TreeNode temp = new TreeNode(root1.val + root2.val);
        temp.left = search(root1.left, root2.left);
        temp.right = search(root1.right, root2.right);
        return temp;
    }
}

700.二叉搜索树中的搜索

二叉搜索树是一个有序树：

• 若它的左子树不空，则左子树上所有结点的值均小于它的根结点的值；
• 若它的右子树不空，则右子树上所有结点的值均大于它的根结点的值；
• 它的左、右子树也分别为二叉搜索树

/**
 * 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;
 *     }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        while(root != null && root.val != val){
            // System.out.println(root.val);
            if (root.val > val) root = root.left;
            else root = root.right;
            
        }
        return root;
    }
}

98.验证二叉搜索树

 

遇到 搜索树，一定想着中序遍历，这样才能利用上特性。 

要知道中序遍历下，输出的二叉搜索树节点的数值是有序序列。

有了这个特性，验证二叉搜索树，就相当于变成了判断一个序列是不是递增的了

• 陷阱1

不能单纯的比较左节点小于中间节点，右节点大于中间节点就完事了。

• 陷阱2

样例中最小节点 可能是int的最小值，如果这样使用最小的int来比较也是不行的。

可以用迭代法模拟二叉树中序遍历，对前中后序迭代法生疏的同学可以看这两篇二叉树：听说递归能做的，栈也能做！ (opens new window)，二叉树：前中后序迭代方式统一写法(opens new window)

/**
 * 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;
 *     }
 * }
 */
class Solution {
    Queue<Integer> list = new LinkedList<Integer> ();
    public boolean isValidBST(TreeNode root) {
        search(root);
        int n = list.size();
        int temp = list.poll();
        while(n > 1){
            int temp2 = list.poll();
            if(temp2 <= temp) return false;
            temp = temp2;
            n--;
        }
        return true;
    }
    public void search(TreeNode node){
        if (node == null) return;
        search(node.left);
        list.offer(node.val);
        System.out.println(node.val);
        search(node.right);
    }
    
}