654.最大二叉树
class Solution {public TreeNode constructMaximumBinaryTree(int[] nums) {return constructMaximumBinaryTree1(nums,0,nums.length);}public TreeNode constructMaximumBinaryTree1(int[] nums,int leftIndex,int rightIndex){if(rightIndex - leftIndex < 1){ //区间没有元素return null;}//区间只有一个元素 直接为叶子节点或者根节点if(rightIndex - leftIndex == 1){return new TreeNode(nums[leftIndex]);}//找到nums数组中最大值以及下标 根节点int index = leftIndex;int maxValue = nums[index];for(int i = leftIndex+1;i < rightIndex ;i++){if(nums[i] > maxValue){maxValue = nums[i];index = i;}}TreeNode root = new TreeNode(maxValue);//根据maxIndex划分左右子树root.left = constructMaximumBinaryTree1(nums,leftIndex,index);root.right = constructMaximumBinaryTree1(nums,index+1,rightIndex);return root;}
}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 = root1.val + root2.val;root1.left = mergeTrees(root1.left,root2.left);root1.right = mergeTrees(root1.right,root2.right);return root1;}
}
队列迭代 只操作root1树 所以无需考虑root2树中节点为空root1不为空的情况 如果为空 直接使用root1值
class Solution {public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {if(root1 == null) return root2;if(root2 == null) return root1;Queue<TreeNode> que = new LinkedList<>();que.add(root1);que.add(root2);while(!que.isEmpty()){TreeNode node1 = que.poll();TreeNode node2 = que.poll();node1.val = node1.val + node2.val;//如果两树左节点都不为空if(node1.left != null && node2.left != null){que.add(node1.left);que.add(node2.left);}//如果两树右节点都不为空if(node1.right != null && node2.right != null){que.add(node1.right);que.add(node2.right);}if(node1.left == null && node2.left != null){node1.left = node2.left;}if(node1.right == null && node2.right != null){node1.right = node2.right;}}return root1;}
}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
这就决定了，二叉搜索树，递归遍历和迭代遍历和普通二叉树都不一样。在搜索值时无需想普通二叉树前中后序全部遍历，以及无需回溯
递归代码 利用二叉树搜索特点
class Solution {public TreeNode searchBST(TreeNode root, int val) {if(root == null || root.val == val){return root;}TreeNode resNode = null;if(root.val > val){resNode = searchBST(root.left,val);}if(root.val < val){resNode = searchBST(root.right,val);}return resNode;}
}递归代码：普通二叉树
class Solution {public TreeNode searchBST(TreeNode root, int val) {if(root == null || root.val == val){return root;}TreeNode left = searchBST(root.left,val);if(left != null){return left;}return searchBST(root.right,val);}
}迭代代码
普通迭代：
class Solution {public TreeNode searchBST(TreeNode root, int val) {while(root != null){if(val < root.val){root = root.left;}else if(root.val < val){root = root.right;}else{return root;}}return null;}
}栈迭代
class Solution {// 迭代，普通二叉树public TreeNode searchBST(TreeNode root, int val) {if (root == null || root.val == val) {return root;}Stack<TreeNode> stack = new Stack<>();stack.push(root);while (!stack.isEmpty()) {TreeNode pop = stack.pop();if (pop.val == val) {return pop;}if (pop.right != null) {stack.push(pop.right);}if (pop.left != null) {stack.push(pop.left);}}return null;}
}队列迭代+搜索二叉树特性
class Solution {public TreeNode searchBST(TreeNode root, int val) {if(root == null){return root;}Queue<TreeNode> que = new LinkedList<>();TreeNode res = null;que.add(root);while(!que.isEmpty()){int size = que.size();for(int i = 0;i < size; i++){TreeNode node = que.poll();if(node.val > val && node.left != null){que.add(node.left);}else if(node.val < val && node.right != null){que.add(node.right);}else if(node.val == val){res = node;}}}return res;}
}98.验证二叉搜索树
思路一：将二叉树中序遍历完保存一个数组 如果这个数组是从小到大 那么这就是一个搜索二叉树，缺点：增加了空间消耗多创建一个数组来存储节点元素
class Solution {public boolean isValidBST(TreeNode root) {List<Integer> arry = new ArrayList<>();inorder(root,arry);for(int i = 1; i < arry.size();i++){//遍历前一位与后一位进行比较 防止数组越界if(arry.get(i) <= arry.get(i-1)){return false;}}return true;}private void inorder(TreeNode root,List<Integer> arry){if(root == null){return;}inorder(root.left,arry);arry.add(root.val);inorder(root.right,arry);}
}思路二：在遍历中就进行大小比较，初始一个节点前面的值 相当于中序遍历 前一个节点与后一个节点连续比较，缺点：如果数据精度跟小就要变化
class Solution {//后台测试数据中有int最小值 只有比int最小值更小才能更新pre的值private long pre = Long.MIN_VALUE;public boolean isValidBST(TreeNode root) {if(root == null){return true;}//左boolean left = isValidBST(root.left);if(!left){return false;}//中序遍历，验证遍历的元素是不是从小到大if(pre < root.val){pre = root.val;}else{return false;}boolean right = isValidBST(root.right);return left && right;}
}思路三：双指针 前一个节点与后一个节点进行比较
class Solution {//用来记录一个前节点TreeNode pre = null;public boolean isValidBST(TreeNode root) {if(root == null){return true;}boolean left = isValidBST(root.left);if(pre != null && pre.val >= root.val){return false;}pre = root;boolean right = isValidBST(root.right);return left && right;}
}
