513.找树左下角的值
题目链接：513.找树左下角的值
文档讲解︰代码随想录(programmercarl.com)
视频讲解︰找树左下角的值
日期：2024-09-12
想法：1.迭代：用层序遍历，遍历每层时记录下第一个节点的值，到最后一层就是要求的值；2.递归：根据最大的深度来找目标值。
Java代码如下：
//迭代
class Solution {public int findBottomLeftValue(TreeNode root) {Queue<TreeNode> que = new LinkedList<TreeNode>();if(root != null) que.offer(root);int res = 0;while(!que.isEmpty()){int size = que.size();for(int i = 0; i < size; i++){TreeNode node = que.poll();if(i == 0){res = node.val;}if(node.left != null){que.offer(node.left);}if(node.right != null){que.offer(node.right);}}}return res;}
}
//递归
class Solution {private int maxDeep = -1;private int value = 0;public int findBottomLeftValue(TreeNode root) {findLeftValue(root,0);return value;}private void findLeftValue (TreeNode root,int deep) {if (root.left == null && root.right == null) {if (deep > maxDeep) {value = root.val;maxDeep = deep;}return;}if (root.left != null) {deep++;findLeftValue(root.left,deep);deep--;}if (root.right != null) {deep++;findLeftValue(root.right,deep);deep--;}return;}
}
112. 路径总和
题目链接：112. 路径总和
文档讲解︰代码随想录(programmercarl.com)
视频讲解︰路径总和
日期：2024-09-12
想法：用减到0来判断。
Java代码如下：
class Solution {private boolean travel(TreeNode root, int Sum){if(root.left == null && root.right == null && Sum == 0) return Sum == 0;if(root.left != null){Sum -= root.left.val;if(travel(root.left, Sum)) return true;Sum += root.left.val;}if(root.right != null){Sum -= root.right.val;if(travel(root.right, Sum)) return true;Sum += root.right.val;}return false;}public boolean hasPathSum(TreeNode root, int targetSum) {if(root == null){return false;}return travel(root, targetSum - root.val);}
}
总结：回溯很晕，还需要点时间。
106.从中序与后序遍历序列构造二叉树
题目链接：106.从中序与后序遍历序列构造二叉树
文档讲解︰代码随想录(programmercarl.com)
视频讲解︰从中序与后序遍历序列构造二叉树
日期：2024-09-12
想法：第一步：如果数组大小为零的话，说明是空节点了；第二步：如果不为空，那么取后序数组最后一个元素作为节点元素；第三步：找到后序数组最后一个元素在中序数组的位置，作为切割点；第四步：切割中序数组，切成中序左数组和中序右数组；第五步：切割后序数组，切成后序左数组和后序右数组；第六步：递归处理左区间和右区间。可以设置中序区间：[inorderBegin, inorderEnd)，后序区间[postorderBegin, postorderEnd)，相当于找到每一层的根。
Java代码如下：
//106.从中序与后序遍历序列构造二叉树
class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {if(postorder.length == 0){return null;}return build(inorder, 0, inorder.length, postorder, 0, postorder.length);}private TreeNode build(int[] inorder, int instart, int inend, int[] postorder, int poststart, int postend){if(postend - poststart == 0){return null;}int rootValue = postorder[postend - 1];TreeNode root = new TreeNode(rootValue);int i = 0;for(; i < inorder.length; i++){if(inorder[i] == rootValue){break;}}int leftinstart = instart;int leftiend = i;//分割点，左闭右开int rightinstart = i + 1;int rightinend = inend;int leftpoststart = poststart;int leftpostend = poststart + (i - leftinstart);//长度等于前序中的int rightpoststart = poststart + (i - leftinstart);//都是左闭右开int rightpostend = postend - 1;root.left = build(inorder, leftinstart, leftiend, postorder, leftpoststart, leftpostend);root.right = build(inorder, rightinstart, rightinend, postorder, rightpoststart, rightpostend);return root;}
}//105. 从前序与中序遍历序列构造二叉树(思路是一样的)
class Solution {public TreeNode buildTree(int[] preorder, int[] inorder) {if(preorder.length == 0){return null;}return build(inorder, 0, inorder.length, preorder, 0, preorder.length);}private TreeNode build(int[] inorder, int instart, int inend, int[] preorder, int prestart, int preend){if(preend - prestart == 0){return null;}int rootValue = preorder[prestart];TreeNode root = new TreeNode(rootValue);int i = 0;for(; i < inorder.length; i++){if(inorder[i] == rootValue){break;}}int leftinstart = instart;int leftinend = i;//分割点，左闭右开int rightinstart = i + 1;int rightinend = inend;int leftprestart = prestart + 1;int leftpreend = prestart + 1 + (i - leftinstart);//长度等于前序中的int rightprestart = prestart + 1 +(i - leftinstart);//都是左闭右开int rightpreend = preend;root.left = build(inorder, leftinstart, leftinend, preorder, leftprestart, leftpreend);root.right = build(inorder, rightinstart, rightinend, preorder, rightprestart, rightpreend);return root;}
}
