Day17--二叉树--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

思路：

前序遍历。寻找子数组的区间。注意区间要统一成习惯。这里是左闭右开。

class Solution {public TreeNode constructMaximumBinaryTree(int[] nums) {return build(nums, findIndexOfMaxValue(nums));}private TreeNode build(int[] nums, int index) {if (nums.length == 0) {return null;}if (nums.length == 1) {return new TreeNode(nums[0]);}TreeNode root = new TreeNode(nums[index]);int[] leftArray = Arrays.copyOfRange(nums, 0, index);int[] rightArray = Arrays.copyOfRange(nums, index + 1, nums.length);root.left = build(leftArray, findIndexOfMaxValue(leftArray));root.right = build(rightArray, findIndexOfMaxValue(rightArray));return root;}private int findIndexOfMaxValue(int[] nums) {int max = Integer.MIN_VALUE;int index = 0;for (int i = 0; i < nums.length; i++) {if (nums[i] > max) {max = nums[i];index = i;}}return index;}
}

思路：

用栈迭代，前序遍历。

把node2的值加到node1上，如果一方是有节点一方是null，创建一个节点赋值为0.

class Solution {public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {// 处理null情况if(root1==null&&root2==null){return null;}if(root1==null){return root2;}if(root2==null){return root1;}Deque<TreeNode> stack = new ArrayDeque<>();stack.push(root2);stack.push(root1);while(!stack.isEmpty()){TreeNode node1 = stack.pop();TreeNode node2 = stack.pop();node1.val += node2.val;if(node1.left==null&&node2.left!=null){node1.left = new TreeNode(0);}else if(node1.left!=null&&node2.left==null){node2.left = new TreeNode(0);}if(node1.left!=null&&node2.left!=null){stack.push(node2.left);stack.push(node1.left);}if(node1.right==null&&node2.right!=null){node1.right = new TreeNode(0);}else if(node1.right!=null&&node2.right==null){node2.right = new TreeNode(0);}if(node1.right!=null&&node2.right!=null){stack.push(node2.right);stack.push(node1.right);}}return root1;}
}

思路：

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

思路：

中序遍历

class Solution {private long prev = Long.MIN_VALUE;public boolean isValidBST(TreeNode root) {if (root == null) {return true;}if (!isValidBST(root.left)) {return false;}if (prev >= root.val) {return false;}prev = root.val;return isValidBST(root.right);}
}