题目一
二叉树题——最大二叉树
根据给定的数组来构建最大二叉树
具体题目如下
解法
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/** * 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 method(nums, 0 ,nums.length- 1 ); } public TreeNode method( int [] nums, int lo, int hi){ if (lo>hi){ return null ; } int index = - 1 ; int max = Integer.MIN_VALUE; for ( int i = lo;i<=hi;i++){ if (max<nums[i]){ max = nums[i]; index = i; } } TreeNode root = new TreeNode(max); root.left = method(nums,lo,index- 1 ); root.right = method(nums,index+ 1 ,hi); return root; } } |
题目二
二叉树题——构造二叉树
根据给定的数组按照指定遍历条件构造二叉树并返回根节点
具体题目如下
解法
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/** * 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 buildTree( int [] preorder, int [] inorder) { return method(preorder, 0 ,preorder.length- 1 ,inorder, 0 ,inorder.length- 1 ); } public TreeNode method( int [] preorder, int preLeft, int preEnd , int [] inorder, int inLeft, int inEnd){ if (preLeft>preEnd){ return null ; } int rootVal = preorder[preLeft]; int index = - 1 ; for ( int i = inLeft;i<=inEnd;i++){ if (rootVal == inorder[i]){ index = i; } } TreeNode root = new TreeNode(rootVal); int leftSize = index - inLeft; root.left = method(preorder,preLeft+ 1 ,leftSize+preLeft,inorder,inLeft,index- 1 ); root.right = method(preorder,leftSize+preLeft+ 1 ,preEnd,inorder,index+ 1 ,inEnd); return root; } } |
题目三
二叉树题——构造二叉树
根据给定的数组按照指定遍历条件构造二叉树并返回根节点
具体题目如下
解法
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/** * 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 buildTree( int [] inorder, int [] postorder) { return build(inorder, 0 ,inorder.length- 1 ,postorder, 0 ,postorder.length- 1 ); } TreeNode build( int [] inorder, int inStart, int inEnd, int [] postorder, int postStart, int postEnd) { if (inStart > inEnd) { return null ; } // root 节点对应的值就是后序遍历数组的最后一个元素 int rootVal = postorder[postEnd]; // rootVal 在中序遍历数组中的索引 int index = 0 ; for ( int i = inStart; i <= inEnd; i++) { if (inorder[i] == rootVal) { index = i; break ; } } // 左子树的节点个数 int leftSize = index - inStart; TreeNode root = new TreeNode(rootVal); // 递归构造左右子树 root.left = build(inorder, inStart, index - 1 ,postorder, postStart, postStart + leftSize - 1 ); root.right = build(inorder, index + 1 , inEnd,postorder, postStart + leftSize, postEnd - 1 ); return root; } } |
题目四
二叉树题——构造二叉树
根据给定的数组按照指定遍历条件构造二叉树并返回
具体题目如下
解法
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/** * 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 constructFromPrePost( int [] preorder, int [] postorder) { return method(preorder, 0 ,preorder.length- 1 ,postorder, 0 ,postorder.length- 1 ); } public TreeNode method( int [] preorder, int preStart, int preEnd, int [] postorder, int postStart, int postEnd){ if (preStart>preEnd){ return null ; } if (preStart==preEnd){ return new TreeNode(preorder[preStart]); } int rootVal = preorder[preStart]; int leftRootVal = preorder[preStart + 1 ]; int index = 0 ; for ( int i = postStart; i < postEnd; i++) { if (postorder[i] == leftRootVal) { index = i; break ; } } TreeNode root = new TreeNode(rootVal); int leftSize = index - postStart + 1 ; root.left = method(preorder, preStart + 1 , preStart + leftSize,postorder, postStart, index); root.right = method(preorder, preStart + leftSize + 1 , preEnd,postorder, index + 1 , postEnd - 1 ); return root; } } |
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原文链接:https://blog.csdn.net/wai_58934/article/details/123054330