随想录刷题笔记 —二叉树篇6 513找树左下角的值 112路径总和 106中序后序构造二叉树

513找树左下角的值

找出该二叉树的 最底层最左边 节点的值

解法一：队列——层次遍历

class Solution {public int findBottomLeftValue(TreeNode root) {Queue<TreeNode> treeNodes = new LinkedList<>();treeNodes.offer(root);int count = 1;int result = root.val;while (!treeNodes.isEmpty()){TreeNode temp = treeNodes.poll();count--;if (temp.left!=null){treeNodes.offer(temp.left);}if (temp.right!=null){treeNodes.offer(temp.right);}if (count==0){if (!treeNodes.isEmpty()){result = treeNodes.peek().val;}count = treeNodes.size();}}return result;}
}

解法二：递归

设置maxdepth变量，记录最大深度。

class Solution {public int findBottomLeftValue(TreeNode root) {fBLValue(root, 1);return resultval;}int maxdepth = 0;int resultval = 0;void fBLValue(TreeNode root, int depth) {if (root.right==null&&root.left==null){if (maxdepth<depth){resultval=root.val;maxdepth=depth;}return;}if (root.left!=null){fBLValue(root.left, depth+1);}if (root.right!=null){fBLValue(root.right, depth+1);}}
}

112路径总和

是否存在 根节点到叶子节点 的路径，这条路径上所有节点值相加等于目标和 targetSum

递归：传入参数value，表示从根节点到此节点的路径值。

当结点是叶子节点时，计算路径值是否=targetSum

class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {if (root==null){return false;}return hasPathS(root, targetSum, 0);}boolean hasPathS(TreeNode root, int targetSum, int value) {if (root==null){return false;}if (root.right==null&&root.left==null){return value + root.val == targetSum;}return hasPathS(root.left, targetSum, value+root.val)||hasPathS(root.right, targetSum, value+root.val);}
}

106中序后序构造二叉树

递归：以后序数组的最后一个元素为切割点，先切中序数组，根据中序数组，反过来再切后序数组。

一层一层切下去，每次后序数组最后一个元素就是节点元素。

class Solution {public TreeNode buildTree(int[] inorder, int[] postorder) {TreeNode root = new TreeNode(postorder[postorder.length-1], null, null);int mid = findindex(inorder, postorder[postorder.length-1], 0, inorder.length-1);if (mid!=0){root.left = buildTree(Arrays.copyOfRange(inorder, 0, mid), Arrays.copyOfRange(postorder, 0, mid));}if (mid!=inorder.length-1){root.right = buildTree(Arrays.copyOfRange(inorder, mid+1, inorder.length), Arrays.copyOfRange(postorder, mid, postorder.length-1));}return root;}int findindex(int[] order, int value, int left, int right){for (int i = left; i <= right; i++) {if (value==order[i]){return i;}}return -1;}
}