力扣513 找树左下角的值

class Solution {public int findBottomLeftValue(TreeNode root) {Queue<TreeNode> que = new LinkedList<>();que.offer(root);int res = 0;while(!que.isEmpty()) {int size = que.size();for(int i = 0; i < size; i++) {TreeNode tmp = que.poll();if(i == 0) res = tmp.val;if(tmp.left != null) que.offer(tmp.left);if(tmp.right != null) que.offer(tmp.right);}}return res;}
}

力扣112: 路径总和

class Solution {public boolean hasPathSum(TreeNode root, int targetSum) {if(root == null) return false;if(root.left == null && root.right == null) return (root.val - targetSum == 0);targetSum -= root.val;if(root.left != null) {boolean left = hasPathSum(root.left, targetSum);if(left) return true;}if(root.right != null) {boolean right = hasPathSum(root.right, targetSum);if(right) return true;}return false;}
}

力扣113: 路径总和II

class Solution {public List<List<Integer>> pathSum(TreeNode root, int targetSum) {List<List<Integer>> res = new ArrayList<>();List<Integer> count = new ArrayList<>();traversalSum(root, targetSum, res, count);return res;}public void traversalSum(TreeNode root, int targetSum, List<List<Integer>> res, List<Integer> count) {if(root == null) return;count.add(root.val);if(root.left == null && root.right == null && root.val == targetSum) {/* 由于 count 是一个引用类型，后续的操作会修改 count，从而影响到已经添加到 res 中的列表。为了解决这个问题，你应该在添加 count 到 res 时创建一个新的列表，以避免后续修改对结果的影响。*/res.add(new ArrayList<>(count));}if(root.left != null) traversalSum(root.left, targetSum - root.val, res, count);if(root.right != null) traversalSum(root.right, targetSum - root.val, res, count);count.remove(count.size() - 1);}
}

力扣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(inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0) {return null;}return traversal(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);}public TreeNode traversal(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {if(postStart > postEnd) return null;TreeNode root = new TreeNode();root.val = postorder[postEnd];if(postStart == postEnd) return root;int index;for(index = inStart; index < inEnd; index++) {if(inorder[index] == root.val) break;}int inStartLeft = inStart;int inEndLeft = index - 1;int inStartRight = index + 1;int inEndRight = inEnd;int postStartLeft = postStart;int postEndLeft = postStart + (index - inStart) - 1;int postStartRight = postEndLeft + 1;int postEndRight = postEnd - 1;root.left = traversal(inorder, inStartLeft, inEndLeft, postorder, postStartLeft, postEndLeft);root.right = traversal(inorder, inStartRight, inEndRight, postorder, postStartRight, postEndRight);return root;}}

力扣105.从前序与中序遍历序列构造二叉树

class Solution {public TreeNode buildTree(int[] preorder, int[] inorder) {if(preorder == null || preorder.length == 0 || inorder == null || inorder.length == 0) {return null;}return traversal(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);}public TreeNode traversal(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {if(preStart > preEnd) return null;TreeNode root = new TreeNode();root.val = preorder[preStart];if(preStart == preEnd) return root;int index;for(index = inStart; index <= inEnd; index++) {if(inorder[index] == root.val) break;}int inStartLeft = inStart;int inEndLeft = index - 1;int inStartRight = index + 1;int inEndRight = inEnd;int preStartLeft = preStart + 1;int preEndLeft = preStartLeft + (index - inStart) - 1;int preStartRight = preEndLeft + 1;int preEndRight = preEnd;root.left = traversal(preorder, preStartLeft, preEndLeft, inorder, inStartLeft, inEndLeft);root.right = traversal(preorder, preStartRight, preEndRight, inorder, inStartRight, inEndRight);return root;}
}