题目:
105
根据一棵树的前序遍历与中序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如,给出
前序遍历 preorder = [3,9,20,15,7] 中序遍历 inorder = [9,3,15,20,7]
返回如下的二叉树:
3 / 9 20 / 15 7
106
根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如,给出
中序遍历 inorder = [9,3,15,20,7] 后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树:
3 / 9 20 / 15 7
思路:
两题思路是相似的,都是通过前序/后序遍历数组确定中间结点,然后找到中间结点在中序遍历的位置。中序遍历序列中,中间结点前的序列就是中间结点的左子树,中间结点后的序列就是中间结点的右子树。按照这样的方法去构建二叉树。
105
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { return build(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1); } private TreeNode build(int[] preorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) { if (preStart > preEnd || inStart > inEnd) return null; TreeNode root = new TreeNode(preorder[preStart]); int leftTreeLength = 0; for (int i = inStart; i <= inEnd; i++) { if (inorder[i] == preorder[preStart]) //找到左子树上结点的数量 break; leftTreeLength++; } root.left = build(preorder, inorder, preStart + 1, preStart + leftTreeLength, inStart, inStart + leftTreeLength - 1); root.right = build(preorder, inorder, preStart + 1 + leftTreeLength, preEnd, inStart + leftTreeLength + 1, inEnd); return root; } }
106
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode buildTree(int[] inorder, int[] postorder) { return build(postorder, inorder, 0, postorder.length - 1, 0, inorder.length - 1); } private TreeNode build(int[] postorder, int[] inorder, int preStart, int preEnd, int inStart, int inEnd) { if (preStart > preEnd || inStart > inEnd) return null; TreeNode root = new TreeNode(postorder[preEnd]); int leftTreeLength = 0; for (int i = inStart; i <= inEnd; i++) { if (inorder[i] == postorder[preEnd]) //找到左子树上结点的数量 break; leftTreeLength++; } root.left = build(postorder, inorder, preStart, preStart + leftTreeLength - 1, inStart, inStart + leftTreeLength - 1); root.right = build(postorder, inorder, preStart + leftTreeLength, preEnd - 1, inStart + leftTreeLength + 1, inEnd); return root; } }