力扣labuladong一刷day30天二叉树

力扣labuladong一刷day30天二叉树

一、654. 最大二叉树

题目链接：https://leetcode.cn/problems/maximum-binary-tree/
思路：采用分解问题的方法，先构造父节点再构造左右子节点，每次都在指定区间内搜索，找到最大值后构造父节点，再划分左右区间递归。

class Solution {public TreeNode constructMaximumBinaryTree(int[] nums) {return traverse(nums, 0, nums.length - 1);}TreeNode traverse(int[] nums, int left ,int right) {if (left > right) return null;int max = -1, index = -1;for (int i = left; i <= right; i++) {if (nums[i] > max) {max = nums[i];index = i;}}TreeNode root = new TreeNode(max);root.left = traverse(nums, left, index-1);root.right = traverse(nums, index+1, right);return root;}
}

二、105. 从前序与中序遍历序列构造二叉树

题目链接：https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/
思路：通过前序和中序构造二叉树，要维护好前序的区间以及中序的区间，先构造父节点再构造左右子节点，父节点是前序区间的left，左右子节点由递归返回。 此外要注意速度，想节省遍历中序数组寻找父节点索引的时间的话，可以使用map预先存储下来。

class Solution {Map<Integer, Integer> map = new HashMap<>();public TreeNode buildTree(int[] preorder, int[] inorder) {for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return create(preorder, inorder, 0, preorder.length-1,0, inorder.length-1);}TreeNode create(int[] preorder, int[] inorder, int left1, int right1, int left2, int right2) {if (left1 > right1 || left2 > right2) return null;int midV = preorder[left1];TreeNode node = new TreeNode(midV);int index = map.get(midV);node.left = create(preorder, inorder, left1+1, left1+index-left2, left2, index-1);node.right = create(preorder, inorder, left1+index-left2+1, right1, index+1, right2);return node;}
}

三、106. 从中序与后序遍历序列构造二叉树

题目链接：https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
思路：本题和上题类似，也是先构造父节点，然后再构造左右子节点，父节点由后续right构造，然后划分中序和后序的区间，递归构造左右子树。

class Solution {Map<Integer, Integer> map = new HashMap<>();public TreeNode buildTree(int[] inorder, int[] postorder) {for (int i = 0; i < inorder.length; i++) {map.put(inorder[i], i);}return create(inorder, postorder, 0, inorder.length-1, 0, postorder.length-1);}TreeNode create(int[] inorder, int[] postorder, int left1, int right1, int left2, int right2) {if (left1 > right1 || left2 > right2) return null;int midV = postorder[right2];int index = map.get(midV);TreeNode node = new TreeNode(midV);node.left = create(inorder, postorder, left1, index-1, left2, left2+index-left1-1);node.right = create(inorder, postorder, index+1, right1, left2+index-left1, right2-1);return node;}
}

四、889. 根据前序和后序遍历构造二叉树

题目链接：https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-postorder-traversal/
思路：通过前序和后序去构造二叉树，构造的结果不唯一，但方法是一样的，每次使用前序的left做为父节点，left+1做为左孩子的值，然后使用这个去后序中去划分区间，然后在划分前序遍历的区间，之后递归构造即可。

class Solution {Map<Integer, Integer> map = new HashMap<>();public TreeNode constructFromPrePost(int[] preorder, int[] postorder) {for (int i = 0; i < postorder.length; i++) {map.put(postorder[i], i);}return create(preorder, postorder, 0, preorder.length-1, 0, postorder.length-1);}TreeNode create(int[] preorder, int[] postorder, int left1, int right1, int left2, int right2) {if (left1>right1 || left2>right2) return null;if (left1 == right1) return new TreeNode(preorder[left1]);int mid = preorder[left1];int index = map.get(preorder[left1+1]);TreeNode node = new TreeNode(mid);node.left = create(preorder, postorder, left1+1, left1 + index-left2+1, left2,index);node.right = create(preorder, postorder, left1 + index-left2+2, right1,index+1, right2-1);return node;}
}